@0celo7 Well..I think there might be horrible morphisms. Are we allowed to assume the morphism is Lie?

@ACuriousMind My proof that $\mathrm{SO}(2)\approx S^1$: First, $\mathrm{O}(2)$ has two connected components inherited from $\mathrm{GL}(2,\Bbb R)$. $\mathrm{SO}(2)$ is the identity-connected component of this, so it has one component. $\mathrm{O}(2)$ is compact, $\mathrm{SO}(2)$ is also compact. (cont. in a bit)

@ACuriousMind We compute the dimension of $\mathrm{SO}(2)$ be looking at its Lie algebra. $\mathfrak{so}(2)\cong\mathfrak{gl}(2)-\mathfrak{sym}(2)\to 2^2-2(2+1)/2=1$. So $\mathrm{SO}(2)$ is a one-dimensional, compact, connected manifold. By the classification theorem, it's a circle.

lol

Lol?

If you knew how a rotation matrix looks, the proof is simply writing down the isomorphism :P

Also, for your proof to work you also need to prove there's only one group structure on the circle

I don't know linear algebra, so that's not happening.

@ACuriousMind Huh?

What group structure?

The $\approx$ is a diffeomorphism of manifolds.

@0celo7 You just showed that $\mathrm{SO}(2)$ is a circle as a manifold

I know.

@0celo7 Oh, but of what use is that if you've a group morphism $\mathbb{R}\to \mathrm{SO}(2)$ that you want to examine?

@ACuriousMind Smooth in $t$? Sure.

@0celo7 You really need to learn linear algebra

@ACuriousMind I took a course, learned nothing, now hate it.

Yup, I'm out :P

So it's really not happening.

@ACuriousMind Hmm.

@Danu There's a bit of QM/intuition discussion a few hours ago in the log if you want something else :P

@ACuriousMind So how do I show that $\mathrm{SO}(2)$ is the circle group?

My proof would be either a physicist proof

Or just to write down the rotation matrix.

@0celo7 Then lust look at the morphism of Lie algebras, which is a continuous linear map $\mathbb{R}\to\mathbb{R}$. Linear maps are $t\mapsto at$ for some $a\in\mathbb{R}$. Exponentiate again to get the Lie group morphism $t\mapsto\exp(2\pi\mathrm{i}at)$.

@ACuriousMind Is exponentiation surjective on abelian Lie groups?

(deja vu)

@0celo7 Well, you solve the condition $AA^T = 1$ for a 2x2 matrix $A$ and deduce you can parametrize the entries as with sin/cos.

then you just write down the isomorphism

@ACuriousMind I tried to do that yesterday, and failed.

I got a lot of shitty equations!

Then tame them!

You know what result you want to get

Exercises don't get much easier than that :P

@0celo7 I don't see why that's relevant.

I couldn't tame my 10 pound cat

But in this case, you can easily check both exponential maps are surjective.

@ACuriousMind Well

@ACuriousMind Is it true in general though

what was I doing

How do I know that every morphism of groups that we reduce to a morphism of algebras can again be lifted to the group

@0celo7 yes

Does that make any sense

@0celo7 Just do it explicitly in this case! Don't bother with Lie theory for 1D Abelian groups

Lie theory is easier than linear algebra, dude

That's really overkill if I ever saw one

@ACuriousMind I can probably come up with something worse.

Spare me.

I was going to troll and say I don't remember how to multiply matrices

But I really don't.

You really need to learn linear algebra

Nah

I remembered

What even is there to learn

@ACuriousMind I'm stupid

If $a^2+b^2=1$, can I suppose $a=\cos\theta,b=\sin\theta$

You can't suppose it, but it is true.

Oh wow, I didn't know that

How do you prove it

Think about it

...

It's probably trivial

I will think about it.

It is

No, it's not trivial

put theta = arctan a/b

@WillO Sure

But what is arctan

Power series I guess

How do you show that the power series trig is geometric trig

Oh dear, here we go down the rabbit hole again...

@ACuriousMind I think that's an important question

It is

So what rabbit hole

Do you know how to see this?

It was also part of my first semester analysis course :P

Hmm

Well I'm not smart

@0celo7 The rabbit hole that this is utterly unrelated to what you were actually trying to do

I guess you use trig stuff to compute trig derivatives

And derive the Taylor series

Whatever

I think we did that in high school

rewrite in polar coordinates, convert a,b to r,theta. your condition is r=1, so theta parametrizes all solns

@WillO I know, but I'm not convinced trig works

I will work on that later

(what even is trig?)

Ok, we must have done this 2x2 orthogonal matrix thing in linear algebra

but I'm missing some concept

the condition just gives me 3 equations

+ det =1 gives me a fourth

@ACuriousMind if $A=[[a,b],[c,d]]$, then I have $a^2+b^2=1,c^2+d^2=1, ac+bd=0$

I'm utterly confused after that

$\begin{pmatrix} a & b\\

hmm

how are you supposed to do a matrix in here

With pmatrix, as you began there

ok, but what do I do with these damn equations

What do you mean "what do I do"?

what are you trying to accomplish?

@ACuriousMind this

I can do it geometrically as @WillO described yesterday

@WillO He's trying to show that $\mathrm{SO}(2)$ is given by the standard rotation matrices

But I don't see an algebraic solution

we did this yesterday

A different way

Then do it the frigging geometric way!

No, I didn't find that very convincing

...

I want to see your algebraic way

(1,0) maps to (x,y) with x^2+y^2=1, so x=cos theta, y=sin theta. etc.

I know

I don't really like that

I'm really losing patience with your "I'm not convinced" attitude. Proofs are proofs. I've got better things to do than to hold your hand through the proof of a statement you already know a proof of.

It wastes my time and yours

@ACuriousMind I'm fascinated that there's an algebraic approach and I want to see it...

I can't figure it out

oh nevermind

I figured it out.

if you are looking to characterize rotations of the plane while pretending you dont know what the circle is, i think you might never be satisfied

@ACuriousMind Sure, thanks.

@ACuriousMind Er, how do you do the inverse caret for Čech cohomology in TeX?

@0celo7 \check in math mode, \v in text mode

Cool

@ACuriousMind It's true in the de Rham case that $H^\bullet_{dR}(M)\cong \check H^\bullet(\mathfrak U,\Bbb R)$, is there such an isomorphism for more general cohomologies than de Rham?

I don't know if that makes any sense

Yes, but you'll have to search for the exact conditions (e.g. the notion of "good cover") yourself because I don't have those memorized. Taking the limit over all covers on the r.h.s. makes the statement true even more generally, but I don't think it holds always

@ACuriousMind Finding good covers on manifolds is fun, I have a full proof on MSE that all manifolds have a good cover (only full proof of that fact I know of)

It's a nice way of showing how Riemannian geometry can be helpful in topology

@ACuriousMind Suppose I have a sequence $0\to A\to B$ of vector bundles, where $A\to B$ is a monomorphism. Why is there a unique (up to isomorphism) vector bundle $C$ such that $0\to A\to B\to C\to 0$ is exact?

The idea is probably to write $C=B/A$ and any other bundle that fits in there must be isomorphic

yes, define B/A fiberwise

then patch in the obvious way

Yeah, but the ismorphism is the issue

@ACuriousMind I was listening to some metal and thought it was pretty good. Turns out it's "Christian rock" lol

This is in any event false in general

What do you mean

Take a non-trivial line bundle on a smooth curve. It has a section (not everywhere non-zero), which amounts to a map from a trivial line bundle.

@ACuriousMind sometimes I hate our generation

@WillO Hmm

the quotient is not a vector bundle

The map should be injective

yes.