Section 10.6 rewrite 3_4_16

Section 10.6 rewrite 3_4_16 PhL 3.4.16 I am assuming the previous 5 sections as they are written up in the original Chapter 10.doc. In this writeup, the basis vectors xei are in the tangent space arrangement. That is xe1, xe2, .... xen span the tangent space at point x on M xen+1 ... xem are all normal to M at point x I threw this idea out, but now I am bringing it back. I will try to trace through this entire section before writing it up. I lean on my discussion in the Shj Ch 6 notes as follows: 1. R = (Dφ) and as usual dx = (Dψ)(t) dt or (Chapter 2). 2. Let {tei} be the usual "e-basis" in t-space, where (tei)j = δij. 3. Write dt = dti (tei) in t-space so that maybe e(t)i for t-space would be better?? dx = Rdt = R[dti (tei)] = dti(R tei) ≡ dti (xei) where xei ≡ R (tei) for i = 1,2..n only The n basis vectors {xe1, xe2..xen} then span the tangent space at x on M. The remaining m-n basis vectors xei for i = n+1 to m in x-space are all normal to M and can be defined in some reasonable manner such that the full set of m vectors {xei} in x-space are linearly independent and form a true basis for x-space = Rm specific to the point x. 4. If tv is a vector in t-space, it transforms just as dx = Rdt transforms, and we have (xv) = R(tv) // in tensor doc had V' = RV ll vectors transform in this manner, for example xei ≡ R (tei) i = 1,2.. n (only) Note that the R matrix is "tall", having m rows and n columns and n < m. 5. Meanwhile we write a general k-form in x-space as αx = Σ'I fI(x) λ^I which in Dirac notation is |αx> = Σ'I fI(x) ( |xei > ^ |xei >....^ |xei > ) = Σ'I fI(x) AltI ( |xei > | xei > .... | xei > ) This differential form is defined on the tangent space to M at point x on M, so each ir ϵ [1,2...k] where k ≤ n. Formally we can write, |αx> ϵ Lk(TxM = xRn xRm) Vk(Rm) Recall that Lk(V) is the space of wedge products of k vectors each of which is an element of vector space V, so Lk(TxM) has V = TxM and we have the space of wedge products of k vectors in TxM. These vectors span a local Rn which we might call xRn and that is embedded in the space xRm obtained by adding the extra m-n basis vectors xei for i = n+1 to m. Any space Lk(V) is part of the larger space Vk(V) spanned by the tensor products of k vectors in V, so here we have Lk(TxM) Vk(TxM) . 6. Recall that R = (Dφ) = a tall m x n matrix. RT is then a wide n x m matrix. Plan A. Consider the following possible threading, |αx> = Σ'I fI(x) ( |xei > ^ |xei >....^ |xei > ) R-1|αx> = Σ'I fI(x) ( |R-1xei > ^ |R-1xei >....^ |R-1xei > ) Then I say xei ≡ R (tei) (tei) = R-1 xei and then the above becomes R-1|αx> = Σ'I fI(x) ( |tei > ^ |tei >....^ |tei > ) I then transpose to get <αx| (R-1)T = Σ'I fI(x) ( <tei | ^ <tei |....^ <tei | ) I then close with a vector set |v1,v2..vk> where these vectors are in t-space <αx| (R-1)T|v1,v2..vk> = Σ'I fI(φ(t)) ( <tei | ^ <tei |....^ <tei | )|v1,v2..vk> = <βt|v1,v2..vk> ?? What I am trying to achieve is this: [φ*(αx)](v1,v2...vk) = αφ(t)(Rv1,Rv2...Rvk) and this does not involve the form βt. Let's try to go backwards from the desired result <φ*(αx)| v1,v2..vk> = <αφ(t)| Rv1,Rv2..Rvk> which I think is really this, where the vi are in t-space, <βt)| tv1,tv2..tvk> = <αφ(t)| R tv1,R tv2..R tvk> Now recall from above that (xv) = R(tv) , so then the above becomes <βt| tv1,tv2..tvk> = <αφ(t)| xv1, xv2, ...xvk> It continues to be a tangled mess! ************************** 7. Next, we define the action of operator RT on our vector |αx> in the space Vk(Rm) according to RT |αx> = Σ'I fI(x) AltI ( RT|xei > RT| xei > .... RT| xei > ) which is basically the idea that Q |v1,v2...vk> ≡ Q|v1> Q|v1> .... Q|vk> . Notice that RT(xei) conforms properly, since xei in Rm has m components. 7. Transposing ** to the dual space the above reads, <αx|R = Σ'I fI(x) AltI ( <xei| R < xei| R .... <xei| R ) Here is the nub of my problem, things always go wrong at this point in my threading. I don't know what to do next. I have been in a Dr Who time loop for the last 3 days on this. Consider: RTxei Then [RTxei]j = Σa=1m (RT)ja (xei)a // regular old matrix/vector multiplication How does this look in the transposed world? [(xei)T R]j = Σa=1m [(xei)T]a Raj = Σa=1m Raj (xei)a These look the same if you can say (RT)ja = Raj which is my old rule in the square matrix world. Suppose I assume this is a correct equation. Then I can write <xei| R = <RT(xei)| = Σa=1m (RT)ja<xei| 7. Now we claim that [R (xei)]a = Σb Rab (xei)b Recall from above that xei ≡ R (tei)