@BalarkaSen I see why it holds

it was not a diffgeo question anymore in the end

we just needed (in the general case) that an injective proper map into a first-countable hausdorff space is homeomorphic to its image

+continuous*

when the domain was compact, it was easy to argue

this more general statement I checked online

that's all we needed right? we already had an injective immersion

to check that it's an embedding, all we needed to do is make a topological argument

at least that's how I see it now

@ShaVukila No no you want the image to be an embedded submanifold

It's not pure topology

A subspace of a smooth manifold which is homeomorphic to a smooth manifold of course need not be an embedded submanifold (in the sense of Lee ie admits slice nbhds)

ye, but if our map is a topological embedding

then since it is assumed to be an injective immersion

it is by definition (lee) an embedding

You're mixing everything up. Why is image of an embedding an embedded submanifold?

That requires a proof

Why not forget these terminologies and imitate my argument above to prove image of proper injective immersions are embedded submanifolds? Mystery solved

Or it could be definition :)

Nooo ted

:(

i know guillemin does that

but lee is conflicting

very confusing

ye, lee is on my side:D

1 sec

I have never read Lee. Despite having known him since he was a grad student.

Just way too wordy for me!

I never liked his smooth manifolds book. It's only good to read some flow proofs from it, and Milnor does it better in Morse theory

Of course, he complained bitterly when he tried teaching out of my undergraduate diff geo notes. He wanted a lot more analytic rigor, and we agreed to disagree on who our audience might be.

Haha

so I don't know what definitions you use Balarka

but Lee defines a smooth embedding as:

an injective smooth immersions (an immersion being a map that has injective differential everywhere) that is also a topological embedding

I don't care about definitions Sha. I am just asking you to prove proper injective immersion has image which is an embedded smooth submanifold

embedded smooth submanifold means locally has slice neighborhoods, like you told me

uh... I don't 'care' either, but we have to work with the same definitions :/

otherwise... there is miscommunication

right?

We all agree on what proper, injective, immersion and an embedded smooth submanifold means right?

pops popcorn and pours martini

So no issues there

@BalarkaSen it seems not, to me

ahhh

idk anymore

I am not saying "embedding" anywhere :)

You did say "embedded."

Not to nitpick, but ...

I said "embedded smooth submanifold"

:)

That's one word

¿¿??

so my false assumption was that the image of an embedding is an embedded submanifold, right?

That requires a proof, is all I am saying, @Sha, if we agree on Lee's definitions on those terms.

And is basically the same thing I am asking you to prove :)

right, okay that was my error

It's all good. You should read Guillemin-Pollack instead of Lee for god's sake!

Sounds like I popped popcorn and poured martini for nothing.

lol ted, what were you expecting:p

lmao

Nah, if you have to work with abstract manifolds, G&P is not at the right level.

I was waiting for fisticuffs!

That's a fair point

Ok Sha read Mo Hirsch

I mean, everyone should learn G&P first and then deal with the abstract chart mess.

Agh. Hirsch has too much analysis for first course.

All that function space stuff for transversality is rough.

I like how in second chapter he proves some globalization theorem for sheaves

and then as a corollary gets transversality

I never read or understood what this mysterious globalization is. Don't think I will ever need it

thanks for the suggestions, I might keep G&P on the side for some more intuition

GP is such a pleasant read

@TedShifrin So the globalization theorem states that if $(F, D)$ is a nontrivial structure functor on $X$ which is continuous and locally extendable then $F(X) \neq \emptyset$

Seems right out of nlab

He proves immersions are dense in $C^r(M, N)$ if $\dim N \geq 2 \dim M$ using this. Sigh.

Of course, I adore sheaves. But I've never worked through that chapter systematically.

Oh right I can't dunk on sheaves in front of a complex geometer.

You can dunk on derived categories, though.

That sucks

Oh good point lol

Soham keeps bothering me with technical stuff at the beginning of G/H to which I've never paid attention (like their way of defining local intersection number in the singular case). I suppose I should take a day to sort it out for him, but I haven't.

OK, I'm out for now. Sleep, Balarka!

Yeah I guess I will

Talk later @Ted, @ShaVuklia

See ya !

bya!

btw @BalarkaSen I see the (your) full argument now, for proper inj immersions that is

cannot thank you enough!

it's way too late, but I finally have some peace x)

wait, does this say that the image of an embedding yields an embedded submanifold?

I think we should only check that the induced smooth structure of the boundary (since in our case, we have an embedding from a manifold to the boundary of sth that we want to prove is a manifold) is comptabile with the one from the manifold

The charts they choose are $(F(U),\phi\circ F^{-1})$, where $(U,\phi)$ is a smooth chart for $N$

so... shouldn't we turn those charts on the boundary into "full" boundary charts?

in the argument that you gave, @BalarkaSen, you show that a "full" chart $V$ of $\mathbb R^2$ turned out to be a slice chart (of the curve), and hence it was a boundary chart for the bounded interior+curve

oh wait, it doesn't matter how the smooth structure looks on the boundary.. all we care about is to find a slice chart, which we have by virtue of the boundary being an embedding submanifold

okay, then I think we have two proofs?

they do assume M has no boundary, but that is fine in our case

I meant N (the domain of the embedding)

@BalarkaSen I was asleep

Mornin'

Good morning Ed

Morning everyone

yo

Did you chose Heidelberg Univeristy over other choices in England ?

yis

I’m asking because I’m myself of the view that education system are quite good in Germany, Netherlands, Switzerland than USA, UK

The UK is a sinking ship and I wanted off that island asap

UK universities are still very good but the country is a pit

How bad is UK?

Ah

It's just

a massive holding pen for rich people to hunt poor people

@EdwardEvans University in UK and USA require very high non-sensical merits

hä

They want us to be excelled in every field of life, I don’t think that should be the criterion

@EdwardEvans like every capitalist country lol?

lol

Anyway the country is being pulled apart at the seams by the Tories and I don't like it so I threw a little tantrum and f*cked off to Germany

I don’t like the way people compete in USA, UK

Everyone should focus on their own knowledge, way of thinking rather than competing

it's probably the same everywhere, just the British have some weird god complex because we UsEd To HaVe A bIg EmPiRe

anyway, gotta go, Algebra awaits

@EdwardEvans ANT?

Algebra 2 lol

what is that

Homological algebra, category theory, and commutative algebra

lol

in imperial, algebra 2 was "prove that C[X] is a PID"

Strong

Leaky I think you needed to qualify a very competitive exam for getting into Imperial College.

it's called the Maths Admissions Test

Does STEP still exist?

Haven't seen it since I left school lol

STEP is for cambridge

Well it was for Oxbridge and Warwick

and google told me it does still exist, idk why I asked lol

@LeakyNun C[X] = ring of real-valued continuous functions on X?

$\Bbb C[X]$

polynomials with complex coefficients

what you said would be $C(X)$

or $C^0(X)$

Yeah, and thats not a PID, right?

well it can be

X is something useful.

X = [0,1]

hmm, what's the easiest proof that C[0,1] is not a PID

I guess maybe just prove that {f | f(0) = 0} isn't principal

a generator would have to be non-zero outside 0

then I don't know what to do next

let's say f is a generator

then let's build a function g such that g(0) = 0 but lim g(x)/f(x) = infty

or just doesn't exist

I think g(x) = f(x) sin(1/x) works

@feynhat there you go

Yup

thanks for your approval

lol

Anyone familiar with Bernoulli differential equations ?

Let $X$ be a topological space, $G$ a topological group that acts continuously on $X$. Why is the map $X \to X/G$ open?

Let $q$ be the quotient map.

If $U$ is open in $X$, then $q(U) = \{\mathcal{O}(x) | x \in U\}$

where $\mathcal{O}(x)$ is the orbit of $x$.

What is $q^{-1}(\mathcal{O}(x))$? Is it same as $\mathcal{O}(x)$?

@feynhat $\mathcal O(x)\subseteq X$, how are you taking $q^{-1}$ of it?

Oh no wait I see what you mean

that notation though

@AlessandroCodenotti Also, $\mathcal{O}(x) \in X/G$.

Alright let's see

Take an open $U$ in $X$, we want to show that $q(U)$ is open

Yes, we need to show $q^{-1}(q(U))$ is open.

But $q(U)$ is open iff $q^{-1}(q(U))$ is open in $X$, right?

Yes. Quotient topology.

Yeah and that's the union of $gU$ as $g$ varies in $G$ because of the orbit thing you were saying

And every $gU$ is open because the action is continuous

(note that the fact that $G$ is a topological group is irrelevant, the important thing is that the action is continuous)

Okay. On the left-hand side, $\mathcal{O}(x)$ is a set and on the other side its an element.

@AlessandroCodenotti I wouldn't know what you mean by action of non-groups.

no I meant that the "topological" in front of group is irrelevant, sorry that wasn't clear

oh right.

@LeakyNun Were you a prodigy ?

no

Did you put some extra effort particularly for that exam?

well i did past papers

just like what i would do before every exam

BTW, this is true, right?

yes

And you qualified it ? Wow!

That's how you get $q^{-1}(q(U))=\bigcup_{g\in G} gU$

Did you solve Olympiad books also for that test?

I did olympiad maths before

not for that test

Means you were a competitive guy from early on

Great. Now onto manifolds. $M \to M/G$. I want to put a smooth structure on $M/G$. By the previous result, I can pushforward the charts to $M/G$.

@feynhat no restriction on the group action?

@feynhat You need some extra assumptions here

$M = \Bbb R^2$, $G = C_2 = \langle \sigma \mid \sigma^2=1 \rangle$, $\sigma(x,y) = (-x,-y)$

Ohh... wait. Yes. That properly discontinuous nonsense.

it isn't nonsense

Still Leaky's action above is properly discontinuous

(As is any action by a finite group trivially)

did you find some differences between studying for completion and studying for the sake of study ?

@AlessandroCodenotti it isn't

I don't know what those terms mean

What's your definition of properly discontinuous?

https://ncatlab.org/nlab/show/properly+discontinuous+action

Mine is "for all compact $K\subseteq X$, the set $\{g\in G\mid gK\cap K\neq\varnothing\}$ is finite"

let's ask what @feynhat's definition is

@LeakyNun What is $M/G$ here though? I mean I can choose a representative in the upper half-plane, but it isn't the upper half plane?

Uh what the nlab calls properly discontinuous I've always seen called wandering

the problem is the "ramification" at (0,0)

@LeakyNun Every point has a neighborhood U, such that $gU \cap U \ne \varnothing \implies g = e$.

@AlessandroCodenotti maybe you mixed up the two terms: https://math.stackexchange.com/q/1719232/328173

@feynhat ok I think that is sufficient

so now you can put an atlas on M/G

Let $\{U_i, \phi_i\}$ be an atlas for $M$.

I would like to choose charts on $M/G$ as $q(U_i)$.

But I don't know how I will define the coordinate.

No,

Okay.

Every point x in M has a neighborhood $V_x$ such that for $g\ne e$, $gV_x \cap V_x = \varnothing$.

So, on this $V_x$, $q$ is invertible, right?

For $\mathcal{O}(x)$ in $M/G$, I choose the chart as $q(U_x \cap V_x)$, and the coordinate map as $\phi_x \circ q^{-1}$?

@LeakyNun

@feynhat sounds reasonable

$q$ is invertible on $V_x$ because, for any $y \in V_x$, $q^{-1}(\mathcal{O}(y))$ will intersect $V_x$ exactly once at $y$. The other members of $q^{-1}(\mathcal{O}(y))$, that is, $gy$ will lie in the corresponding $gV_x$.

I hate the term properly discontinuous. You want the group to be a Lie group and you want the action to be proper. When the group is discrete that gives the condition named 'properly discontinuous'.

@MikeMiller Hi!

But nothing fails to be continuous, it's just continuous for the discrete topology.

Hello

But anyway, the case where G is positive dimensional has a slightly different character to the argument, so it's good to do the discrete case first

@feynhat Also note that q is an open map, and open cts bijections are homeomorphisms.

Did I kill your discussion?

What? No. I thought we were done?

I wasn't really paying attention, I just saw that you stopped. My bad.

Is there a common shorthand for a tuple in which the ith position is m_i and the rest 0?

$m_i \cdot e_i$?

Thats fair but i haven't defined an e_i yet lol

Might call it m_i hat or smth

Uh

$\iota_i(m_i)$ with an implicit understanding of $\iota_i$ being the corresponding inclusion map

What are they tuples of

Elements of an R module

@Thorgott ive just applied iota, which is why i wanna write it explicitly lol

well if you're trying to write it explicitly just write $(0, \cdots, m_i, \cdots, 0)$, where the only nonzero term is the $i$th term, $m_i$

Yeah i guess that's fine, was just wondering if there's a shorthand hehe

Thanks anyway

You're defining the shorthand aren't you

Maybe... Lol, it's just for my own notes so i don't see why not hehe

Just proving that Hom(A,_) commutes with the direct product, and similär for direct sum

Just don't write the formula down above you don't have to

Write $\iota_i$ for inclusion into the $i$th factor of a direct sum and $p_j$ for projection onto the $j$th factor of a direct product and use those

It's clear what those are from the verbal description

Isn't Hom(A, M) = M? *isomorphism

His ring is R. I assume A is some other module.

Okay haha, i think ive mixed up what maps are supposed to be killing each other so i'm gonna start from the beginning; I've already proven it for the direct product, just the direct sum to go lol

Should be basically the same but, as i said, i've mixed up my maps lol

I don't really understand what you mean by mixed up your maps but I suppose I will leave you to it

Dont worry hahaha, just confused myself so i'll start again from the beginning rofl

Thanks

@MikeMiller Oh. I missed that.

A is a reasonable name for an algebra, so I understand the assumption.

Yeah i'm using M, dunno why I wrote A here

@EdwardEvans don't kill each other

Alright I gotta air my confusion: I have a $\varphi \in \operatorname{Hom}_R(\bigoplus_i M_i, M)$ and I send this to $(\varphi \circ \iota_i) \in \prod_{i} \operatorname{Hom}_R(M_i, M)$. Then this should be an isomorphism of abelian groups. That it's a homomorphism is fine, what's confusing me is that the map $$(\varphi \circ \iota_i) \mapsto \bigoplus_i (\varphi \circ \iota_i) : (m_i) \mapsto \sum_i (\varphi \circ \iota_i)(m_i)$$ should be the inverse of the guy above

I think..

right, and $\sum_i (\varphi \circ \iota_i)(m_i) = \sum_i \varphi((0,\dots, 0, m_i, 0, \dots))$.. is that now the same thing as $\varphi((m_i))$?

rip notation

oh

lol

being dumb, got it

@ShaVuklia Whatever you said earlier looks correct to me

@Alessandro Yeah no issue I was just going to pass some point set topology crap through you. It's fine

w h a t

Lmfao

Beautiful

> "one hour left xD"

haha

that made my day

There are other questions too

SteezO-

what a legend

incredible posting

Why do you take this so seriously ? it's the internet, no one cares except for you

Woah

only geeks like you care, gonna report you too haha

What a comeback

@SteezO-

I am dying man

These guys are too good

Hshaha

the exam must be almost over by now, im worried :/

Wish we could find where they're taking their exam and report, top banter

watch out geek, he's gonna report you too

haha

Fight me nöörd

Bluff them; say you found their institution and see how fast the questions get deleted

Is it possible to get banned from MSE ?

definitely

you can get IP banned

Oh wow

That's good

This is $so(1,1)=\{2\times2~ \mathscr{matrices }~X|e^{tX}\in SO(1,1)~ \forall t\}$

not sure if they are both $so(1,1)$ or the first definition is another beast.

guys, what does it mean that I've listened to the same song in the past 3 weeks? the count is now 84.5 hours

maybe corona is catching up anyways

Lol that happens often

if we have 3 equation with 3 variables can we write it as a function of two variable?

equations

what sort of equation

non linear

write what as function of two variable

shouldn't you expect a discrete set of solutions?

x=-st/2+s, y=t+s, u=t+s/2

write u(x,y)

that's not 3 variables

that's 5 variables

correct sorry

3 eq, 5 variables

@Semiclassical I got the answer $\frac{F}{2m} ~(b^2 -4)$ for that question. Is my answer correct? You said that your Lagrangian method could serve as a good check, that’s why I have asked you.

I just wrote my first numerical derivative in fortran.

It took awhile I had to get the epsilon delta definition down.

I have a ring $R$ and an $R$-module $M$. Suddenly this book starts talking about $M$ as a $\Bbb Z$ module as well

So I guess what that means is to have $am=\varphi(a) m$ for $a\in\Bbb Z$ and $m\in M$, where $\varphi\colon \Bbb Z\to R$?

I have no clue, but that sounds like a natural interpretation

@AlessandroCodenotti any R-module is an abelian group

any abelian group is a Z-module

Oh derp

Well that's the same as what I said

Yeah Alessandro that's correct

Z is the initial object in the category of commutative unital rings

So you get a functor from R-Mod to Z-Mod

That is the natural change of structure

The point is that whenever I have a morphism of rings $\varphi\colon S\to R$ I can turn an $R$-module into an $S$-module as I did above in the special case $S=\Bbb Z$ then

adjoint reacc only

A morphism R -> S induces a functor S-Mod -> R-Mod, that is correct.

Don't say any comma category over $S$ nonsense

Any category is naturally a slice category over it's initial object

Sure

And products are just fibered products over the final object

And we're just restating things in a less clear way :P

I thought that's what category theorists do

You like this right

This is what you wanted

Scum

@BalarkaSen Are you implying that we are category theorists or?

you're definitely a category theorist

you cant claim not to be after thinking about compact objects in locally presentable categories

or whatever the shit it was

But that was kinda interesting

what about cocone diagrams

a group homomorphism is a bijection because of a cocone diagram

Also I discovered in the meantime that Vopenka's principle can be phrased in terms of presentable categories even though I don't know why anybody would do that

Nuts

Do you want me to ask why someone would want to do that

no thanks

Nice