The h Bar

ACuriousMind

23:00

@0celo7 Because the horizontal/vertical notion requires a bit more of a geometrical apparatus than just defining the connection forms and covariant derivatives by hand.

0celo7

@ACuriousMind but why is it sometimes done

(that previous comment did not answer my question)

ACuriousMind

@0celo7 I, personally, have no intuition for connection forms and covariant derivatives. I can live with the pictures in Riemannian geometry, but in the general case, I don't see what they mean geometrically (and I suspect many others feel the same). However, a connection form is equivalent to an assignment of horizontal spaces, and the notion of vertical space is natural. And for the horizontal/vertical notion, the geometrical meaning is far clearer.

It tells you e.g. WTF it is that curvature measures - the failure of the Lie bracket of two horizontal vectors to be horizontal.

While just writing $\mathrm{d}\omega -[\omega\wedge\omega]$ doesn't tell you at all why curvature would be of geometric interest

0celo7

so one can treat principal bundles and their curvature without using horziontal/vertical stuff?

ACuriousMind

@0celo7 Well, you can just define the connection form without motivation and define the curvature by the Cartan structure equation (which is what is very often indeed done)

So, yeah, you can do it without ever mentioning vertical and horizontal vectors, but I find that that is precisely where the geometry lies

And, besides, it is the notion of verticality and horizontality that generalizes to arbitrary smooth bundles, and then subsumes all other kinds of connections.

So, from the abstract viewpoint, it is the "correct" definition.

0celo7

ok

The way that Jost does it is he treats general bundles. Then he justifies the Riemannian stuff by looking at sectional curvature.

So he back-justifies the generalized stuff.

ACuriousMind

23:11

Well, how you want to do it depends heavily whether or not you assume the reader is already familiar with e.g. Riemannian curvature, I think

Or, for instance, if you want to teach a physicist, it will certainly be quicker to never talk about the horizontal stuff and instead take the gauge covariant derivative as sole motivation

0celo7

Jost confuses me. It's supposed to be an intro but he proves Hopf-Rinow on like page 17.

do Carmo takes his sweet time getting there.

@ACuriousMind Yeah...

Apparently the best book for connections is still Kobayashi-Nomizu.

ACuriousMind

@0celo7 lol. Hopf-Rinow was like the second-to-last lecture of the class on differential goemetry I took

0celo7

Page 35.

Still the first chapter.

@ACuriousMind Now that I've got you going on connections. What's the difference between the good ol' spin connection and the thin Straumann defines on page 649?

I feel like we've been over this, but it still confuses me.

The spin connection is the connection on the frame bundle, right?

ACuriousMind

@0celo7 I think everything I know about the spin connection is in this answer.

If your answer isn't in there, I don't know it, I never had to work with a spin connection thus far, so I never had to learn more about it

0celo7

@ACuriousMind I edited that thing :P

John Duffield

23:19

@ACuriousMind : Sounds to me as if he's unhappy about some answers he'd had elsewhere. I'll deal with it.

ACuriousMind

@0celo7 Yes, in August. Given your memory lately, it is not at all certain you recall any of it ;P

0celo7

Wait!

ACuriousMind

What have I done.

0celo7

Wtf is the Christoffel symbol, anyway

Is it a connection form?

ACuriousMind

That I've said multiple times by now, I think - it is a $\mathfrak{gl}(n)$-valued connection form on the tangent bundle.

0celo7

23:21

Ok so the tangent bundle is the associated of some $\mathrm{SO}(n)$ bundle

the frame bundle?

ACuriousMind

Yeah, something like that, I don't wnat to say something that's wrong

0celo7

@ACuriousMind so wtf is the spin connection

ACuriousMind

@0celo7 It's only $\mathfrak{so}(n)$-valued (or $\mathfrak{so}(1,n-1)$ in the Lorentzian case)

0celo7

@ACuriousMind I know that!

so we somehow "restrict" the Christoffel in some weird frame

ACuriousMind

I think it has something to do with reduction of the structure group, and you want to reduce to $\mathfrak{so}$ to be able to talk about spinors.

0celo7

23:24

I think Wald and Straumann are doing the same things

but Wald's terribad notation is obscuring the truth

ACuriousMind

The notation in differential geometry is generally horrible, yes.

0celo7

Wald's notation: index free notation is too clumsy and coordinate notation is not precise, so let's add more indices than coordinate notation and call it good

wonder if BBS has any insight

BBS is a paragon of rigor and clarity after all

ACuriousMind

@0celo7 Ha. Ha. Ha.

0celo7

@ACuriousMind You're the fool who bought it.

ACuriousMind

@0celo7 Don't remind me :(

0celo7

23:28

nope, nothing

next book...BLT

MGA

Sorry for the novice question, but I couldn't find the right keywords to google: in Faraday's law, if the loop does not lie in a plane, which is the surface that matters in the calculation of the induced EMF?

0celo7

never apologize for novice questions

@ACuriousMind I'm confused, what's a connection

BLT is unhelpful as well

ACuriousMind

@MGA You can choose any surface whose boundary is the loop.

user54412

@MGA Any surface with the loop as a boundary works

user54412

This is also true in the case where the loop is in a plane

ACuriousMind

23:32

@0celo7 A Lie algebra-valued 1-form that transforms by the usual transformation law for a connection form/gauge field

user54412

the surface doesn't have to be in the plane

MGA

Thanks guys, that's what I thought (because the surface is not uniquely defined!)

0celo7

@ACuriousMind that was a joke

MGA

Is there a named theorem behind this?

ACuriousMind

@MGA Stokes' theorem.

0celo7

23:33

as in BBS has confused me so much I have no clue what's going on

MGA

@ACu

user54412

@MGA The trick is the integral is of B dot dA -- some surfaces lend themselves to this calculation better than others

MGA

@ACuriousMind Stokes or Kelvin-Stokes?

0celo7

you reading Wiki

MGA

Yes, Kelvin-Stokes seems like complicated stuff :-)

ACuriousMind

23:34

@MGA Uhhh...I tend to call all of them just Stokes'...

user54412

I think Green's is the special case here

user54412

Stokes' theorem

In vector calculus, and more generally differential geometry, Stokes' theorem (also called the generalized Stokes' theorem) is a statement about the integration of differential forms on manifolds, which both simplifies and generalizes several theorems from vector calculus. Stokes' theorem says that the integral of a differential form ω over the boundary of some orientable manifold Ω is equal to the integral of its exterior derivative dω over the whole of Ω, i.e., This modern form of Stokes' theorem is a vast generalization of a classical result. Lord Kelvin communicated it to George Stokes in...

MGA

Excellent, thanks a lot guys, that clears it up

I clearly need to properly learn vector calculus

user54412

@MGA The thing is, vector calculus has a lot of "integral of a derivative = integral of a thing on the boundary" and it turns out these are all special cases of one theorem that's too advanced to be taught at that level.

user54412

Once you learn the fully general Stokes theorem, you won't even bother with all those specific cases. Certainly you won't bother to remember them.

ACuriousMind

23:38

I've always thought that the proof of the generalized Stokes' theorem is too difficult compared to how simple the statement is and how easily one can see it in low dimensions graphically.

But searching for a "simple proof" gives...nothing one could teach in analysis or vector calculus.

MGA

But if I study undergrad-level vector calculus it should become clear to me why the surface does not matter, right?

user54412

Perhaps?

MGA

Cool

user54412

Is it clear to you why the integral from a to b of f(x) depends only on F(a) and F(b) (the antiderivative)?

MGA

Yes

Is Stokes' basically just that for surfaces?

user54412

23:41

the fundamental theorem of calculus is just yet another special form of stokes

ACuriousMind

@MGA Yes, the fundamental theorem of calculus is also just a special case of the generalized Stokes' theorem.

0celo7

@ACuriousMind let's see

MGA

@ChrisWhite When I say it's clear, I mean I know that's the fund. thm. of calculus, and I can visualize it by considering that whichever way the function goes, it must eventually go the other way

user54412

@ACuriousMind The way we repeat each other -- it's like we're confused socks or something

0celo7

there are totally simple proofs

MGA

23:42

@ChrisWhite Is there a better insight than that which I'm not aware of?

0celo7

proofwiki.org/wiki/General_Stokes'_Theorem

ACuriousMind

@ChrisWhite Varying lag in the connection of our master to us, probably :P

user54412

@MGA It's hard to say what counts as "better" when it comes to math intuition.

MGA

True, but yes, I do have an intuition for the FTC

It's cool to learn that it's a special case of the generalized Stoke's theorem

Would a text like Spivak's Calculus on Manifolds be the right place to learn about this rigorously?

user54412

@0celo7 I want you to find the nearest vector calculus class (not one in Germany), and use the phrase "finite family of relatively compact charts"

0celo7

23:46

@ChrisWhite probably one in Zurich

ah!

that's what we need!

@ACuriousMind $\mathcal{C}$ is relatively compact

would that work?

user54412

@MGA From the title, yes, though I haven't actually read the book.

ACuriousMind

@0celo7 We?

@0celo7 would what work?

MGA

Thanks again, that was extremely helpful.

0celo7

@ACuriousMind Yes?

@ACuriousMind $\mathcal{C}$ being relatively compact

ACuriousMind

@0celo7 That's a property, not a procedure that can "work". What do you want to deduce from the fact that it is relatively compact?

And what is $\mathcal{C}$, anyway?

0celo7

23:54

@ACuriousMind The thingie we cut from $\Sigma$

ACuriousMind

Are you talking about that asymptotic flatness?

0celo7

@ACuriousMind Yes

ACuriousMind

If you're talking about the ball in Schwarzschild $\mathbb{R}^3-\{0\}$, it's not relatively compact either. Note that my first proposal to fix the definition was that $\mathcal{C}$ be homeomorphic to a relatively compact subsets of $\mathbb{R}^n$ (I just didn't use that word)

0celo7

why is it not relatively compact

ACuriousMind

Because its closure is itself?

Relatively compact is only different from compact if your set isn't closed.

0celo7

23:59

@ACuriousMind Whoops. Forgot the origin is literally not in the manifold :(

Wait how do we actually know it's not a compact set?

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