confusion about the meaning of (omega)star

Confusion about the meaning of ω* PhL 6.29.15 I was trying to understand things in terms of tensor doc, and this led to confusions. My first two conjectures as to the meaning of ω* both failed to give correct results on BOTH pages 409 and 411, but the third conjecture seems to have worked in both these cases. Here is the original text taken from Buck Ch 7 raw notes, just to archive this confusion: Digression on page 409. What happens to differential forms under a transformation? ω ≡ A1dx1+A2dx2 = A(x) dx x = (x,y) x' = (u,v) ω' ≡ A'1dx'1+A'2dx'2 = A'(x') dx' // using tensor doc notation But this is NOT the kind of transformation they are talking about. They instead want you to think of ω* ≡ A'1dx1+A'2dx2 = A'(x') dx (*) // my conjecture of ω* meaning [Note 1: whereas on page 401 we had a 1-form ω going to a 2-form ω*, here we have a 1-form ω going to a 1-form ω* (clearly stated as being a 1-form). I conclude that this is a new and different meaning for the symbol ω* . But I have to decode how here they get from ω to ω*, it is not clear.] Then we can use dx' = Rdx dx = Sdx' to get ω* ≡ A'(x') [Sdx'] = A'i [Sdx']i = A'i Sijdx'j = [A'1 S11 + A'2 S21] dx'1 + [A'1 S12 + A'2 S22] dx'2 = [A' S11 + B'S21] du + [A' S12 + B' S22] dv Now from tensor doc, So then [ see p 409 A] S11 = ∂ux = φ1 S21 = ∂uy = ψ1 S12 = ∂vx = φ2 S22 = ∂vy = ψ2 Then you get ω* = [A' φ1 + B' ψ1] du + [A' φ2 + B' ψ2] dv = page 409 E Since the above expression agrees with p 409 E (assuming A* means A' etc), I think my conjecture (*) above is correct. OK, so we are on the same page (I think). Go back then to ω ≡ A(x) dx x = (x,y) x' = (u,v) ω* = A'(x') dx ω* = A'(x') dx as simple example, So they are defining ω* by transforming only the k-form coefficients, not the differentials. This is like the active rotation of a vector which acquires new components in the same coordinate system. They are allowed to define a "transformation of a k-form" any way they want, and so ω* is what they want to do. Now one benefit of this definition of ω* is the claim d(ω*) = (dω)* They show this with some examples, but maybe I can prove it more generally: ω ≡ A(x) dx dω = d[A(x)] dx ω* = A'(x') dx d(ω*) = d[A'(x')] dx Before doing more, let's work on these two results dA(xi) = ∂jA dxj or dAk(xi) = ∂jAk dxj Then since dω = d[A(x)] dx = d[Ak(x)]dxk = ∂jAk dxj dxk dω = ∂jAk dxjdxk = [ ∂jAk dxj] dxk = [ (∂jA) dxj] dx = - [(Ak)dxk] dx // two ways way #1 way #2 (dω)* = [ (∂'jA') dx'j] dx = - [('A'k)dx'k] dx // two ways way #1 way #2 Meanwhile, d[A'(x')] = ∂'jA' dx'j or dA'k(x'i) = ∂'jA'k dx'j Then way #1 d(ω*) = ∂'jA'k dx'j dxk = [ (∂'jA') dx'j] dx = (dω)* as shown above !!! So I have then proven the little theorem: Theorem PL1: if ω ≡ A(x) dx and ω* = A'(x') dx, then d(ω*) = (dω)*. Comment: That fact that I get the result d(ω*) = (dω)* lends further credence to my conjecture (*) above about the "meaning of symbol ω* ". Theorem PL2: Now consider: ∫γ ω* = ∫γ [A'(x'(x)) dx ] // my interpretation of ω* as above = ∫γ' [A'(x') dx' ] // run x' along γ' rather than x along γ = ∫γ' [A(x) dx ] // dot product is scalar in En if A is a vector = ∫γ' ω // my interpretation of 1-form ω = ∫γ* ω // their name for γ' curve. The above is then my proof of (7-31) A. [ but who says A is a vector ? ] Theorem PL3: Next, consider ∫∫D (dω)* = ∫∫D [ (∂'jA'(x'(x)) dx'j(x)] dx // one of my "two interp ways" above = ∫∫D' [ (∂'jA'(x')) dx'j] dx' // run dx'jdx'k over D' instead of D = ∫∫D' [ (∂jA(x)) dxj] dx // dot product is scalar = ∫∫D' (dω) // my interp above = ∫∫D* dω // their name for D' The above is then my proof of (7-31) B. But there are some subtleties here! Consider V'(x') = (∂'jA'(x')) dx'j V'k(x') = (∂'jA'k(x')) dx'j = (∂'jA'k)dx'j = component k of a vector in x'-space So our integrand is [ (∂'jA'(x')) dx'j] dx' = V'k(x') dx'k = a true scalar for any transformation! = V'(x') dx' = V(x) dx // since a true scalar This occurs in the first proof as well where we have [A'(x') dx' ] = A'k dx'k but in order to really see this, you have to use covariant notation with up and down indices. I can see why Buck's don't want to get into that right here. Does Arapura address this stuff? The answer is NO. I would have to find some other source. I just found a pretty good source that talks about more advanced things like pullbacks and I think that does relate to the above, but let's not wander off right now!! Status: I will go with my little derivations of page 410 A and B (7-31). These were hugely non-obvious and Bucks are a little dishonest in their presentation I would say. But let's now move on. The proof of Theorem 7 is now clear: Theorem 7: If Green's Theorem holds in x-space, it also holds in x'-space. Note that this is for an arbitrary transformation they call T, and I call F. Perhaps this would have some implication for curvilinear coordinates. So far we don't have much motivation for why Theorem 7 is important. Theorem 8: Proof that (dω)* = d(ω*) This is a 3-page proof. But I think I did my own proof above as Theorem PL1. I wonder if there is any connection between our proofs? Mine is probably no good. My proof was only for 1-forms! 0-forms. ω = f(x) dx = Sdx' contravariant ω* = f '(x') = f(x(x')) // seems only possibility // same as page 411 A d[f(x(x'))] = ∂if dxi = ∂ 'if ' dx'i // your choice, both ways are right d(ω*) = d[f(x(x'))] = ∂if dxi = f dx = f [Sdx'] // probably what B says Let's try to explicitly confirm that the above agrees with p 411 B: the following based on p 411 F: S11 = ∂ux = φ1 S21 = ∂uy = ψ1 S31 = ∂uz = θ1 S12 = ∂vx = φ2 S22 = ∂vy = ψ2 S32 = ∂vz = θ2 Then d(ω*) = f [Sdx'] = Σi=13∂if Σj=12Sij dx'j = Σi=13fi Σj=12Sij dx'j = f1 [S11dx'1 + S12dx'2] + f2 [S21dx'1 + S22dx'2] + f3 [S31dx'1 + S32dx'2] = [f1S11 + f2S21 + f3S31] dx'1 + [f1S12 + f2S22 + f3S32] dx'2 = [f1φ1 + f2ψ1 + f3θ1] du + [f1φ2 + f2ψ2 + f3θ2] dv and this DOES agree with p 411 B. We now continue: dω = df = f dx = this is a 1-form // this is what C says (dω)* = 'f'(x') dx // according to my 1-form treatment above I know that f '(x') = f(x(x')) so then 'f'(x') = ' f(x(x')) and then ['f'(x')]i = ∂ 'i f(x(x')) = fj ∂xj/∂x'i = fj Sji and finally, (dω)* = 'f'(x') dx = ['f'(x')]i dxi = fj Sji dxi = f1(S11dx1 + S12dx2 ) + f2(S21dx1 + S22dx2 ) + f3(S31dx1 + S32dx2 ) = f1(φ1dx1 + φ2dx2 ) + f2(ψ1dx1 + ψ2dx2 ) + f3(θ1dx1 + θ2dx2 ) This looks like the first line of D, but they have dx1' and dx2' whereas I have dx1 and dx2. Something is wrong. We would agree if I used (dω)* = 'f'(x') dx' instead of (dω)* = 'f'(x') dx, but then this conflicts with my initial conjecture (*) above! Let's back up and try again from p 411 C: dω = df = f dx = ∂if dxi = fi dxi (dω)* = ∂'if' dxi = ∂f '/ ∂x'i dxi = df/∂xj * ∂xj/∂x'i dxi = fj ∂xj/∂x'i dxi = fj Sji dxi = fj [Sdx]j = fj [SSdx']j = fj [S2 dx']j = f1 [ (S2)11 dx'1 + (S2)12 dx'2] + etc = a big mess which is just plain wrong. So I am unable to get from C to D on page 411. Note that dx = φ1du + φ2dv So in line D, Bucks are simply rewriting things with x,y,z just "replaced" with their x(u,v) and so-on functions. That is not my ω* conjecture! Let's now back up to that conjecture and reexamine it. Here is the conjecture (*) which I made: ω ≡ A1dx1+A2dx2 = A(x) dx x = (x,y) x' = (u,v) ω* ≡ A'1dx1+A'2dx2 = A'(x') dx (*) // my conjecture of ω* meaning But let's now consider a different conjecture for the meaning of ω*. ω ≡ A1(x)dx1+A2(x)dx2 = A(x) dx x = (x,y) x' = (u,v) ω* ≡ A1(x(x') )dx1'+ A2(x(x') )dx2' = A(x(x')) dx' ≡ A1*(x')dx1'+ A2*(x')dx2' = A*(x') dx' (**) This then is a different meaning for A1*. We have A1*(x') ≡ A1(x(x')) This is not the same as tensor doc's thing A1'(x') ≡ R1j Aj(x) for a vector field Now, let's try out conjecture (**) on page 409: ω ≡ A1dx1+A2dx2 = A(x) dx ω* ≡ A1*(x') dx'1+A2*(x')dx'2 = A*(x') dx' = A1*(u,v) du + A2*(u,v) dv But this disagrees with p 409 C. OK, let's now try a third conjecture: ω ≡ A1(x)dx1+A2(x)dx2 = A(x) dx x = (x,y) x' = (u,v) ω* ≡ A1(x(x'))dx1+ A2(x(x'))dx2 = A(x(x')) dx ≡ A1*(x')dx1+ A2*(x')dx2 = A*(x') dx (***) This agrees with p 409 C. Now jump ahead to page 411 and see what conjecture (***) says there: dω = df = f dx = this is a 1-form // this is what C says Now let's define g(x) ≡ f(x) gi(x) = ∂if(x) g*(x') ≡ g(x(x')) g*i(x') = gi(x(x')) This is a little tricky for stupid me, so try not to screw it up please: g(x) ≡ [f](x) g*(x') = [f](x(x')) = [f]*(x') It is fighting me hard. How do you "write this out"? Strong mental block in progress. [f](x(x')) = just express [f](x) in terms of x' coordinates! [∂if](x(x')) = just express [∂if](x) in terms of x' coordinates! ∂if = ∂f/∂xi = fi(x(x')) dω = df = f dx = ∂if dxi = fi(x) dxi (dω)* = fi(x(x')) dxi = fi dxi = fi [Sdx']i = fiSijdx'j = f1[S11dx'1 + S12dx'2] + 2 more terms = f1[φ1dx'1 + φ2dx'2] + 2 more terms = f1[φ1du + φ2dv] + 2 more terms = agrees with D (finally). So now I need to push this confusion off to a separate doc and just do the correct version in these raw notes. Recall from tensor doc and dx = Sdx' Therefore (dω)* = [STf(x) ] [Sdx' ] = [STf(x) ]i[Sdx' ]i = [STik∂kf(x) ][Sijdx'j] = [Ski ∂kf(x) ][Sijdx'j] = [SkiSij] ∂kf(x) ][dx'j] = wrong = = 'f'(x') [Sdx'] // does NOT agree with D I need to review this * operator going back to page 408 where it is first used.