It's 5:00 now, I'll see whether after sleeping I can get my brain clearer on this. I am interested in the results of exercise 5 and 6 and I also believe it will be very important for the later sections

3 apples grated

Only 19 to go

I should have spruced for a mechanical grater

@BalarkaSen What's wrong with it?

Important physics work

That looks disgusting

Oxidized apple pulp does look gross

@Slereah I met a french girl yesterday. She had nice teeth

what an odd thing to notice

Do people have really awful teeth in general where you live

@Slereah Not really, I just like good teeth

@0celo7 If I had accepted that, I'd have to accept "let $(M, (U_\alpha, \varphi_\alpha), (TM, M, T_xM \cong \Bbb R^k, GL_n(\Bbb R^k)), (J:TM \to TM, J^2 = -I))$ be an almost complex manifold"

@Slereah Maybe her teeth were the best thing about her? oO

Maybe she was just a disembodied set of teeth floating above the ground

Maybe she was a mimir

Were my profile picture still Morte, I'd make the obligatory lewd remark :P

@ACuriousMind knows what's up

@BalarkaSen nah. (M, S, TM, J) is good enough.

S for smooth structure.

@ACuriousMind Good call on the protection of the sunset question.

I deleted the non-answers out of hand, but left the others to be buried in a hail of downvotes and deleted by community action.

@0celo7 It is very hard to make jokes that are both dry and subtle work over a text channel. Dry and over the top can work, as can subtle but pointed. Trying to have both adjectives at the same time is asking for trouble.

Heck, it is tough to make the combination work in person, when you get right down to it.

@dmckee I enjoy being stupid.

It wasn't meant as a joke.

By the same user. Could be a legit question, but it will probably turn into a crap show

Yes. Someone trying to avoid getting hit by the q- and a-bans right away, I suppose.

Isn't calculating e.g. how far one can see from the top of a ship until the curvature blocks your sight a completely standard exercise for applying the Pythagorean theorem?

@ACuriousMind Yep.

I'm not sure if I am impressed by somebody gaming the system just to keep posting nonsensical answers. I guess he/she actually believes in what he/she is saying enough to want to keep saying it

::sigh::

The "young artist" bit sounds intriguing. I bet whatever help is given, if any, will end up as a diagram in one of the youtube videos or facebook group

@ACuriousMind o.o

How?

How do you do that on a sphere

Christ

Where do I learn about this?

Also, not that it matters much for 10km or 20km... but it probably needs to be clarified if it's 10km in a straight line, or 10km arc length...

@dmckee Please take down the starred message calling me a nutcase.

It's highly inappropriate coming from a room owner.

The clutching construction is nice. Need to read that carefully tomorrow.

Clutching construction?

Isomorphism classes of oriented k-dimensional vector bundles on S^n are in 1-1 correspondence with maps S^{n-1} --> GL_k^+(R).

@BalarkaSen Do you mean homotopy classes of maps?

Yeah, thanks.

I could have sworn we've talked about that before, in the $S^1$ case.

I think we have.

::resists temptation to star the message asking to have the "nutcase" message unstarred::

@dmckee Your steadfastness is an inspiration to us all.

only 9 apples left

Four hours for 10 apples?!

Ok, let's see

@ACuriousMind Got the best grade on the topology test, bet the TAs are wishing they let me in their circle of fun now

Doesn't look as if your production scheme will scale up well, @Slereah

So we have a principal bundle

Well if I make something nice maybe I'll invest in some better tools

@ACuriousMind He's a madman spending hours to make alcohol. Why doesn't he go to the store like any other terrorist

Because reaping the fruits (or drinks) of your own labour has its own sweet taste?

yes

Plus you can force people to drink it

"Here try it, it's my own!"

Can you?

Maybe it's just urine

Trust me I'm 99% sure there is no urine

@Slereah Whether that works or not depends on how you're known to people

You don't want to be known as a guy who pees in jars

They might think you'll mix them up

Like, I definitely wouldn't have taken anything from that dude in my school who put a pack of apple juice on the heating and left it there for weeks to "ferment" it.

I also wouldn't take anything from the crazy GR nut.

Does it even work with apple juice packs

I'd guess whatever additives they put in it kills the yeast

@Slereah No, the stuff just began to cultivate a nice mold.

The smell was horrific

Whoops, had my name in it.

@ACuriousMind Do you know anything about Radon-Nikodym derivatives?

I only know of Krusty's radon detector

@0celo7 No.

Typo.

Apparently Gaussian curvature is a Radon-Nikodym derivative.

I shall investigate this.

But I am still stuck on principal bundle fun times

@ACuriousMind The elven lord wore a full 3-piece suit with a pocket watch to class today

Nice

Howdy

I don't remember, am I ignoring you?

It looks like Trump is going insane

@0celo7 Who? Me?

@0celo7 What makes you say this now?

because I didn't think so before?

Okay, why do you think so now?

he was tweeting insults at 3AM

it's the 3AM part that bothers me

@0celo7 I'd say the more troubling thing is that he tweeted this at 12:20am.

I don't see anything wrong with that

@0celo7 When would you read a physics paper without sources?

I don't read physics papers.

0celo7 wouldn't read physics papers

Heh

@ACuriousMind Ok, I am thoroughly confused by associated bundles. So we have the homeo $\pi^{-1}(U)\approx U\times G$ from the principal bundle. How the heck do we get $\pi^{-1}_E(U)\approx U\times F$ in the associated bundles?

Everyone pretty much says this is trivial, what am I missing?

Let me check if there's something in my eternally unfinished writeups

I've verified that they are isomorphic as sets

And I think I have continuity

But the continuity of the inverse is eluding me

@ACuriousMind Can you please take a look at KN Vol. 1 page 54? They claim it follows from the action of $G$ on $\pi^{-1}(U)\times F$.

I understand how that action works, but not how it implies the result

Just had a chat with Dave Bacon, one of the Scirate dudes.

He likes the idea of adding an issue tracker.

Will be discussing the idea with the devs soon. There are a couple technical issues to discuss.

@0celo7 Well, you define the projection $\pi_V : P\times_G V\to M$ from the associated bundle just by $\pi_V(p,v) = \pi(p)$ for $\pi : P\to M$ the projection of the principal bundle and checking it's well-defined on equivalence classes, right? On a trivializing $U$, you have that $\pi_V^{-1}(U) = \pi^{-1}(U)\times V/{\sim} = U\times V$ just by thinking about the equivalence relation. Since both $\pi^{-1}$ and quotienting out the relation are continuous, so is $\pi_V^{-1}$.

@DanielSank Interesting. Did you have any prior involvement with them or did you just "walk up" to them and suggest that?

whoa whoa whoa ACM, slow down

Why is $\pi^{-1}$ continuous?

And how does that give me a homeomorphism?

@SirCumference I wonder how he'd feel about applying the same standard to his remarks?

@0celo7 Ah, that was badly formulated, it isn't...

@dmckee Everything Trump says is correct

But it's not what you want, anyway

Trump said so

@ACuriousMind Well, wait.

$\pi$ is a projection map $P\to M$.

So it's open.

therefore it must be true

By definition of quotient topology

So yeah, $\pi^{-1}$ is continuous?

The two equalities I wrote are homeos, though, and you wanted the homeo $\pi_V^{-1}(U) = U\times V$.

@0celo7 Nah, it's not a proper function

Taking the inverse image isn't a function on the actual spaces

Oh, oh, right.

@ACuriousMind I can't figure either of those out!

Well, the first is just writing out the definition of what $\pi_V^{-1}$ is.

So is that an equality or a homeo?

And the second is by $\pi^{-1}(U) = U\times G$ and $U\times G\times V/{\sim} = U\times V$.

@0celo7 It's a true equality

ahhh

@ACuriousMind why $U\times G\times V/\sim=U\times V$

that's exactly what I need

Well, on $P\times V$, the relation is $(p,v)\sim (q,w) \iff \exists g\in G: q = pg \land w = \rho(g)^{-1}v$. On $U\times G\times V$, this means that $(x,g,v)\sim (y,h,w)\iff \exists k\in G: x=y\land h = gk \land w = \rho(k)^{-1}v$.

Since the action is free and transitive, to each $(x,g,v)$, $(x,1,\rho(g)v)$ is the only representant of its class with $(x,1)$ in the first two entries, and so specifying $x$ and $w\in V$ suffices to fix the equivalence class and no two such choices belong to the same class, hence the quotient is $U\times V$.

Is that a \land?

All "wedges" are logical ands, yes

you know how I feel about logical quantifiers :/

That's not a quantifier

Quantifiers are $\forall,\exists$.

you know what I mean

What is $\rho$?

The representation of $G$ on $V$. I'm thinking of associated vector bundles, but you can replace it by a more general action on the fiber without any problems, I think

Ah, ok

(reading now)

@ACuriousMind Is? You mean homeomorphic?

I am perfectly able to argue that $\pi^{-1}_E(U)=U\times V$ as sets.

@ACuriousMind Walked up and suggested.

There's some history though.

I used github to write my most recent paper. It was a very positive experience.

The issue tracker allowed us to divvy up the work and assign things.

Git itself meant that we coauthors could work in parallel in a reasonable way.

The issue tracker was a superstar though. It made revisions so much easier.