As I said, look at a real-valued function, $y=f(x)$. Why is $f^*dy = df$? For me, that is the definition.

@Balarka: I worked it all out. I think the OP didn't mean torsion in the technical sense. But I have a (messy) answer to his question, relating the constant angle to the length of one turn around the torus.

ok putting that aside, to ensure that I get the next step right, $F^\ast \mathrm d\omega = F^\ast (\mathrm d\omega_I \land \mathrm dx^{i_1} \land \cdots \mathrm dx^{i_k}) = F^\ast \mathrm d\omega_I \land F^\ast \mathrm dx^{i_1} \land \cdots F^\ast \mathrm dx^{i_k}$ $= \mathrm dF^\ast \omega_I \land \mathrm dF^\ast x^{i_1} \land \cdots \mathrm dF^\ast x^{i_k} = \mathrm d (F^\ast \omega_I \land F^\ast x^{i_1} \land \cdots \land F^\ast x^{i_k}) = \mathrm d F^\ast \omega$

wait what am I doing

hi italic @Alessandro

No, you messed up, but you started right, @Leaky :)

$F^\ast \mathrm d\omega = F^\ast (\mathrm d\omega_I \land \mathrm dx^I) = F^\ast \mathrm d\omega_I \land F^\ast \mathrm dx^I = \mathrm dF^\ast \omega_I \land F^\ast \mathrm dx^I = \mathrm d((F^\ast \omega_I) (F^\ast \mathrm dx^I)) = \mathrm dF^\ast(\omega_I \mathrm dx^I) = \mathrm dF^\ast\omega$

the trick is to ignore the rest of the terms :P

@TedShifrin also today I noticed a subtlety in the proof of Stokes' theorem

You do need to use $d^2=0$.

@LeakyNun how bizzare

Which subtlety, Leaky?

so usually we would expand $\int_M \mathrm d\omega = \int_M \rho_i \mathrm d\omega$ which would not be very successful; instead one needs to expand $\omega = \rho_i \omega$ and then $\int_M \mathrm d\omega = \int_M \mathrm d (\rho_i \omega)$ to reduce it to the local case; indeed Spivak pointed out that $\mathrm d(\rho_i \omega) = \rho_i \mathrm d\omega + (\mathrm d\rho_i) \land (\mathrm d\omega) = \rho_i \mathrm d\omega + (\mathrm d(1)) \land (\mathrm d\omega) = \rho_i \mathrm d\omega + 0$

You need sums when there isn't Einstein in play. You can't just write one index.

sorry

@anakhro sure, OK!

No, you don't actually need to do that. Spivak made a mess of that.

indeed, one can avoid that by starting from the other side, i.e. expanding $\int_{\partial M} \omega$

but hey wasn't he your teacher... :P

Yes, but you can just immediately argue that both sides are linear in $\omega$ and reduce to a single $\rho_i\omega$.

I did it that way in my book/lectures, but Spivak's book goes through all that mess that's unnecessary.

indeed

it was quite subtle for me

did Spivak also go through that mess when teaching you?

I honestly don't remember. I should look at his big Diff Geo book.

Yeah, he does the same thing ...

well I just extracted it out of that book, so...

are we talking about the same book?

Oh, I thought you were looking at Calculus on Manifolds.

and I still have no idea at all what exterior derivative should mean geometrically

wasn't this concept developed rather early in the history of maths

and concepts developed earlier tend to have more meaning

I guess it's about determinants being volumes of parallelpipeds

Differential forms came along surprisingly late ... turn of the 19th/20th century or so.

That's nothing to do with d. That's what exterior algebra is about ... determinants.

like why on earth should d(fω) = df ∧ dω

like I can "explain" df = f_i dx^i using "infinitesimal changes"

Huh? That was totally wrong.

which one?

"Why on earth"?

oh sorry I meant d(fω) = df ∧ ω

Still no good.

Only true if $\omega$ is closed.

The product rule is geometric: It's about areas of rectangles ...

ok then why on earth should d(fω) = df ∧ ω + f dω

but ω isn't a 0-form

Huh?

You can reduce by linearity over functions to the case of $f dx^I$.

ok then why should d(fdx^I) = df ∧ dx^I?

so d should be interpreted as "infinitesimal element of"?

I certainly don't say that. But the way to understand that is by thinking of Stokes's Theorem on a little box.

The author of the book I'm working through is attempting to conclude that $S_6$ is not a complete group, and the author is doing this by noting that $Aut(S_6)/Inn(S_6) \simeq C_2$. I can see that this implies $Inn(S_6) \neq Aut (S_6)$, but I am wondering if there is a quick way to see that $Aut(S_6)/Inn(S_6) \simeq C_2$.

Just work in two dimensions and understand $d(f\,dy) = \frac{\partial f}{\partial x}\,dx\wedge dy$.

so I should think about the vector field correpsonding to $f~\mathrm dy$?

Arnolds interpretation is using the taylor expansion of the volume of a box using stokes

i think in classical mech

good book

Yeah just read Arnold

I would just integrate over a little rectangle and see that Stokes's Theorem dictates what $d$ should mean.

I mean, we have to stipulate something.

I agree with this

i.e. find $\mathrm d$ such that $\int_D \mathrm d(f~\mathrm dy) = \int_C f~\mathrm dy$

for $D$ a sufficiently small rectangle and $C$ its boundary?

Right.

Doesn't need to be small, but that's helpful for intuition localizing the definition.

ok so the bottom and top edge give 0

the right hand side becomes $f(x,0) - f(0,0)$?

for a rectangle with bottom left corner at origin, x-span being x, y-span being 1?

I think this is a really bad estimation

ok let the y-span be y and let f be relatively constant along the y-direction

so the right hand side becomes $y(f(x,0) - f(0,0))$

Huh? Left-hand side is $f(0,0)-f(0,y)$ and right-hand side is $f(x,y)-f(x,0)$?

oh...

I meant the right hand side as in the right hand side of the equation

and then $y(f(x,0) - f(0,0)) \approx xy f_x$

hey this actually works

Well, what I wrote is nonsense, but I was following your nonsense :P

There you go, that's the right idea ...

well if $f$ is constant along the $y$-direction then it isn't nonsense

I wrote $\Delta f$, which is nonsense.

I am chatting too many places. :P

so those "verify the stokes theorem for the rectangle" exercises I did in multivariable calculus... are useful??

or rather, the intuition is still, "approximate by small rectangles"

Yes, approximate by small rectangles to get intuition.

and for non-top forms, I just take a closed submanifold of appropriate dimension?

Or a parallelepiped, yeah.

ok so just let $f = F_y$, and then the right hand side of the equation becomes $F(x,y) - F(x,0) + F(0,0) - F(0,y) \approx y(F_y(x,0) - F_y(0,0)) = y(f(x,0) - f(0,0)) \approx xy f_x$

hey I can't believe it

hey

what are you guys talking about?

I recall the $f=F_x$ trick helping me prove the symmetry of partial derivatives

@topologicalmagician Stokes theorem (generalized for manifolds)

ahh, that's interesting

If you don't like the usual mean value theorem proof, I recommend Fubini's Theorem.

how does one show that the set $\{$ (x,y,z): f(x,y,z)=c $\}$ is a manifold in R^3? What function satisfies the homeomorphism condition?

@topologicalmagician implicit function theorem

You can't do that except locally, in general, by using the implicit function theorem. Go watch some of my lectures :P

@LeakyNun Thanks!

@TedShifrin hey!, I was watching your lectures all night

hey isn't it also orientable @Ted

LOL, I'm so sorry, @topologicalmagician.

huh? @Leaky

the level set

@TedShifrin all is forgiven :P, I enjoy them, you're really great at explaining concepts.

Well, if you can argue it's a compact surface without boundary, then, yes, @Leaky.

@Ted or I can just point its normal to the direction grad f?

If it's everywhere nonzero ... What about the infinite Möbius strip in $\Bbb R^3$, Leaky?

is it a level set? hmm

Can you write it as the zero-set of a smooth function?