Mathematics - 2017-11-29 (page 6 of 6)

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11:00 PM

So... how do we interpret this form, though?

Like, given $p \in M$ and $X \in T_pM$, how do we see $\omega_p(X)$ in terms of that section?

@MatheinBoulomenos You have to be careful here, because such things as what Balarka wrote can most certainly interpreted pointwise (i.e. saying that for every point you get a Hom), where the setting is in fact finite-dimensional.

So I'd say such statements still make sense

MatheinBoulomenos

Writing $\Gamma(TM)^* \otimes \Gamma(E)^* \otimes \Gamma(E)$ is not pointwise

@TastyRomeo $\omega_p(X)$ is an endomorphism of $E$

Mathei has a point. I am not sure what the answer is.

I have not thought about this

@Antonios-AlexandrosRobotis Α ναι; Πού έμενες στην Ελλάδα;

what are you doing?

11:03 PM

@MatheinBoulomenos But saying $\Gamma(TM^*\otimes E^*\otimes E)$ is fine, no?

MatheinBoulomenos

@Danu yes

Ok, so Balarka was not careful in the way he wrote it

But everybody knows what is meant, I guess :P

@MikeMiller Who are you asking? :P

the five people who appear to be in one conversation where they’re doing something

haha

Well, I only know the very basic definitions of vector bundles

11:05 PM

Balarka was trying to help TastyRomeo understand something about covariant derivatives I think

MatheinBoulomenos

but wait, don't we need the maps from $\Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ to be tensorial (i.e. $C^\infty(M)$-linear) in both components? This means that we can take the tensor product over $C^\infty(M)$

So initially when I had a morphism $\Gamma(T^*M \otimes E) \to \Gamma(E)$, I should have immediately converted that to the space of sections of the bundle of bundle morphisms $T^*M \otimes E \to E$?

And I want to find a one-to-one correspondence between covariant derivatives and one-forms for rank-1 vector bundles

@MikeMiller Mathei is poking a hole in my loose argument to show that the difference of two connections is an End(E)-valued 1-form

@TastyRomeo so you need (1) cov. der's form an affine space over End-valued oneforms

(2) the End-bundle of a line bundle is trivial

MatheinBoulomenos

11:06 PM

@BalarkaSen your argument is fine, it's just your notation that isn't

((3) there always exists a cov der on a line bundle)

Hm, I see.

(3) is what I need to take a canonical cov der to take the difference with, right?

Yes

Yeah

you don't need it to be canonical

11:07 PM

An origin in the space of connections, so to say

there is in general no canonical choice

Well, I mean, I choose it to be the canonical one

There is no canonical one

But it doesn't matter

you just pick one

After he chooses it is a canonical one :P

hahaha

11:08 PM

The choice isn't canonical is the point

That's what I meant, but yeah, whatever :P

but yeah

he won't recognize it next time he comes back :D

true

The end-bundle stuff I should be able to manage.

11:08 PM

(in some special cases you do have a canonical choice)

Yes, you should be able to prove that the End-bundle of a line bundle is trivial

I gave two arguments for that

(and more generally, what do you know of the End-bundle of a bundle of arbitrary rank?)

Not much, since I barely know anything but the definitions.

But I'd wager that it squares the rank of the bundle?

yep

Also that

but from the proof that the End-bundle of a line bundle is trivial

you'll get a related statement for aarbitrary End-bundles

11:12 PM

Hi @MikeM ... long time no see (except for that adorable kitty picture).

I figured I needed a nowhere zero section to get triviality?

But you have one.

And that I would probably get from composing the transition maps

But, yeah, the End(E) stuff I can work out the details I suppose.

Following up on the determinant I wrote earlier: a use of the double-angle identity implies $$f(\cos^2\alpha,\cos^2\beta,\cos^2\gamma)=\frac14 \begin{vmatrix} 1 & -\cos 2\alpha & -\cos 2\beta \\ -\cos 2\alpha & 1 & -\cos 2\gamma \\ -\cos 2\beta & -\cos 2\gamma & 1\end{vmatrix}$$

You can do it "intrinsically". It works for any vector bundle, but getting one nowhere-zero section tells you that for line bundles it's trivial.

11:14 PM

So I guess I can follow that the difference of two covariant derivatives is an End(E)-valued one-form.

@Semiclassic: Should I be giving this as homework to my 9th graders? :)

Which ought to vanish when $\alpha+\beta+\gamma=0$

Lol

@TedShifrin Exactly

(so you'll get a trivial rank-1 subbundle for any End-bundle)

MatheinBoulomenos

@TedShifrin you should absolutely give your 9th graders homework on vector bundles

No, no, @Mathei: second conversation.

11:15 PM

You could, actually

you konw, talk about Moebius strips ;D

We do linear algebra at the end of our course. I'll give it then.

lol @Mathei

You missed my folding-paper-into-the-cube problem that one of the kiddies got in 4 minutes (when I hadn't gotten it in an hour).

@TastyRomeo So what's the nonvanishing section?

user image

So ... fold that into a cube, folding only along the obvious given creases.

11:18 PM

We can't fold diagonally, right? Ugh

NOOOO. Just along the "horizontal" and "vertical" folds.

My "weakest" kid got it ... and was so happy to show it to the "strongest" kids who didn't. Not to mention all the geometers/topologists on Facebook who were stuck.

MatheinBoulomenos

user image

Narcissus jewel

I can attest that there are no 'tricks' and it's a 'legit' solution only folding over the obvious lines that appear. (Took me ages :P )

@MatheinBoulomenos Copyright infringement on my memes.

There is a word that I would give as a hint to someone who was stuck for hours and hours. When I saw what my student did, just casually, I was so angry that I'd missed that idea.

11:20 PM

I am going to sue you for this

Narcissus jewel

B&

MatheinBoulomenos

@BalarkaSen lol

Narcissus jewel

@TedShifrin I think you gave me the hint, but I didn't understand it, until after I'd solved it :P.

Ohhh, I forgot I'd told you.

Probably because you've changed names.

But I've forgotten anyhow.

Narcissus jewel

I was just called 'user0124104' and some random numbers before. I had a purple identicon.

11:22 PM

I need to cut out a paper for this

Yes, Balarka, I did that for me and my students.

I actually love it when students or former students show me up by getting things more cleverly or faster than I do. :)

@Ted Hi

No, that isn't sarcastic.

Me too, as long as they don’t then think I’m just dumb :p

I won't return the hi, because it was a return of my earlier hi :)

11:23 PM

@Danu Ugh, I'm so confused right now. I was going to say "the identity" but that doesn't make sense, does it?

Ayup.

@Tasty It does

Oh...

@MikeM: I doubt that'll happen. Usually your students think you're too smart.

Right, maybe not confused then.

11:24 PM

@TastyRomeo Good :D

Just the map $E \to E$ which sends fiber to fiber by the identity map

That's correct, @Tasty.

And you can do that for any vector bundle.

So any End bundle has a trivial rank 1 subbundle

I was about to set you on fire

Yeah, Danu said that a while ago.

LOL

11:25 PM

@MikeMiller lmao

But you do that lots less than you used to, @MikeM.

im on a loose mode

please spare my pyping

i shouldnt be helping people

You should be folding paper.

I am making my piece of paper squarer

Squares are rather essential.

11:26 PM

i hope it doesn't turn into a rhombus

I used a ruler to fold my paper against.

i have no rulers available for me :(

So End(E) is trivial, meaning that End(E) = M x R ?

yep

Edge of book?

(A virtual book will not suffice.)

11:27 PM

Right, but what I don't get then is, like

$\omega_p(X) \in End(E) = M \times \mathbb{R}$

fucking hell it's shorter on the wider side now

makes no comment

But I think I'm supposed to get a "normal" one-form, with image in $\mathbb{R}$ ?

@TedShifrin i'll use my laptop monitor

this is very low production quality

Lower even than mine was.

11:28 PM

I cut a piece out with a pocket knife... it's close enough to square ^^

How a $1$-form, @Tasty?

I'm surprised you all aren't complaining that you have no piece of paper to use.

pfft

Attempting to be a grumpy old-timer, are ya? :P

Ok, managed to make it right

@TedShifrin I don't understand the question

Oh god

How do I trisect it now

Ah, wait, my browser tabs quadrisect it.

11:31 PM

You said "I'm supposed to get a normal one-form"?

Like $TM \to \mathbb{R}$

Right, but how can a map $E\to E$ be such an animal?

If $E=TM$, a bundle map $TM\to M\times\Bbb R$ is a $1$-form.

@Danu: I've been grumpy as long as you've known me.

wow perfect. i marked out a quadrisection using my browser tabs by putting it on the monitor

I have no idea anymore :(

is this what it feels like to make low fi hip hop?

MatheinBoulomenos

11:33 PM

@BalarkaSen lol

E is just any rank 1 vector bundle $E \to M$. So definitely not $TM$ in general.

@Tasty: Don't worry. It isn't ;P

Right, and you're mapping $E$ to $E$.

There's never a tangent bundle of $M$ appearing.

Wait a second... it's 9 squares but a little cube just needs 6

nods @Danu

Overlaps are allowed.

Indeed, required, since there's that damn hole in there.

Yeah

I feel like this can't be too hard now...

haha

11:36 PM

I am almost done with mine

@TedShifrin Well, not in the endomorphism, no.

Right, @TastyRomeo.

How do I cut the middle square

I have no knife

maybe i can slice it by my pen

Make a hole in the middle

tear towards the folds

right, doing that

11:39 PM

ROFL ... Everyone here should be an engineer. Then all the roads and bridges would fall down.

they already are

Good point.

Given the maths skills of my engineering students I don't think we'd do much worse :^)

Finally done!

But, anyway, right now I have $End(E) = M \times \mathbb{R}$-valued one-forms and I want $\mathbb{R}$-valued one-forms. Do I just cut the $M$-component or so?

11:42 PM

Oh, that's your point. Because $\text{End}(E,E)$ is trivial, an $\text{End}(E,E)$-valued $1$-form is just a regular $1$-form.

You're not understanding the notation.

A trivial-line-bundle-valued $1$-form "is" a regular $1$-form.

You need to understand what a vector-bundle-valued differential form is.

That hole in the middle is annoying

It's sort of the whole point! :P

But I'm fine with annoying you.

@TedShifrin Well, I thought it'd be a one-form that takes values in some vector bundle instead of $\mathbb{R}$, but clearly there's more to it? :P

MatheinBoulomenos

Am I being condescending if I call exercises easy that others don't find easy?

Well, if you understand what you said, yes. So what does it mean to take values in a trivial line bundle?

Yes, @Mathei ... But that's your nature.

11:48 PM

Supa hot fire ohhhhhhh!!!!

Sadly, it's the nature of too many mathematicians/teachers/grad students.

DOnt even need to click play....

nope

heya @Kevin

@Ted I'm thinking of the usual star-like thing you fold to get a cube

the tetris shape

it's not helping

11:49 PM

I'm not sure that's helpful.

LOL

Howdy @Ted

I hesitate to tell you this because it might encourage you to disappear forever, but because you were away there were several things about my homework that I had to figure out for myself

That's great. But Balarka and Mike and PVAL and plenty of others were still here to bug.

I was actually busy with exams

Well, maybe I should disappear.

I did bug Balarka some; he was very helpful

Antonios-Alexandros Robotis

11:51 PM

actually, @TedShifrin Quick qn, if you have a moment.

LOL, yes?

i am only helpful if i am not in my lo-fi mode

which is current what i am

I think you should give up mathematics and just become a rapper, Balarka.

Antonios-Alexandros Robotis

Given diagonal entries on a matrix, can we make the determinant whatever we want by controlling the other entries?

Solved the cube lol

11:52 PM

Yes.

@TedShifrin But I am not a rappa

Antonios-Alexandros Robotis

Thought so

Good for you, @Tasty.

@TastyRomeo come on dude

Antonios-Alexandros Robotis

Pretty sure I've proved it, but I'm literally sick and tired... and my head is spinning.

11:53 PM

@Antonios: As long as you're not requiring the matrix to be upper-triangular :P

Antonios-Alexandros Robotis

lol

that would certainly be a problem.

You can do this in a very boring/easy way.

Antonios-Alexandros Robotis

Yeah, I think I see it.

Hint: Make almost all of 'em $0$.

Antonios-Alexandros Robotis

yeah then you can just abuse cofactor expansion

11:54 PM

Right, back to my actual problem

Or think about block matrices ...

That's a very powerful technique.

@TedShifrin I thought it meant that you got values (p,x) in M x R

Antonios-Alexandros Robotis

hmmm. Cool. I'll be finishing this off then.

Thanks!

@Tasty: But if you're at a point $p\in M$, the values of an $E$-valued $1$-form are in $E_p$, which in our case is $\Bbb R$.

Waiiiiiit

I thought "1-forms on M taking values in End(E)" meant that $\omega_p(X) \in End(E)$

11:57 PM

Sorry, I was too lazy to modify. I just said it for an arbitrary vector bundle $E$. Substitute the correct thing.

No, what you wrote is wrong.

But you're saying that, if you choose the $p$ for the one form, you take the same choice in $E$ ?

$\omega_p(X)\in \text{End}(E_p,E_p)$.

$\omega(X)$ \in End(E)

Oh.

Riiight

In one case $X$ is a tangent vector at $p$, and in the other it's a global vector field, but yeah.

11:59 PM

I would like to say that the proof of Mayer-Vietoris felt like doing Sudoku

just for the record

@Tasty In exchange for my help you will tell me how to do the problem. I am a 14 year old kid so I won't get sued by Epic Games

The cube problem?

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