Testing a claim that phistar=R REVIEWED

Testing the claim that φ* = R PhL 3.17.17 5.17.16: two items here (1) working with the messy t-space to x-space notations like xu . I decided not to show any of this but to relegate it to Appendix E. (2) at this point I am still thinking of φ* as a synonym for R. But later I made two changes to this idea. The first is that R became Dirac operator R, and the second that φ* became a function not an operator. I just had a wrong threading, that's all! I like the final presentation in wedge doc Section 10.7. This is taken from "paradox v1 3_14_16.doc". 3/19/16. Preliminary. I want here to "map" all my new tensor-doc Section 7.19 facts to Picture F. For example, in Section 7.19 I have this major data: (un)i = ui un = <ui | un > = gin = u'i u'n = < u'i | u'n > (en)i = ui en = <ui | en > = Sin = Rni (en')i = e'i e'n = < e'i | e'n > = g'in = ei en = <ei | en > (un')i = e'i u'n = <e'i | u'n> = Rin = Sni (7.19.12) <en | S | e'i> = <e'i | R | en> = g'in <en | S | u'i> = <u'i | R | en> = Sin = Rni <un | S | e'i> = <e'i | R | un> = Rin = Sni <un | S | u'i> = <u'i | R | un> = gin . (7.19.19) The forward rules are these: e' → xe e → te u → tu u' → xu So I will copy the above block, make it red, do edits making things black (tun)i = tui tun = <tui | tun > = gin = u'i u'n = < u'i | u'n > (ten)i = tui ten = <tui | ten > = Sin = Rni (xen)i = xei xen = < xei | xen > = g'in = tei ten = <tei | ten > (xun)i = xei xun = <xei | xun> = Rin = Sni (7.19.12) <ten | S | xei> = <xei | R | ten> = g'in <ten | S | xui> = <xui | R | ten> = Sin = Rni <tun | S | xei> = <xei | R | tun> = Rin = Sni <tun | S | xui> = <xui | R | tun> = gin . (7.19.19) Here then is the resulting data block: (tun)i = tui tun = <tui | tun > = gin = u'i u'n = < u'i | u'n > (ten)i = tui ten = <tui | ten > = Sin = Rni (xen)i = xei xen = < xei | xen > = g'in = tei ten = <tei | ten > (xun)i = xei xun = <xei | xun> = Rin = Sni (7.19.12) <ten | S | xei> = <xei | R | ten> = g'in <ten | S | xui> = <xui | R | ten> = Sin = Rni <tun | S | xei> = <xei | R | tun> = Rin = Sni <tun | S | xui> = <xui | R | tun> = gin . (7.19.19) 2. Want to test the following claim: [ try this in all-Dirac notation sep doc] (a) φ*T = RT and φ* = R [assume] for the pullback matrix. I think this is true from two sources: I derive it in "is phi star symmetric v2" It comes out at the end in "Section 10_6 v2". Now consider my definition of φ*T as stated in Ch 10 v 3 (b) φ*T(xui) ≡ Σj=1n Rij (tuj) i = 1..m [ok] (10.6.9) I claim that in Picture F' we have (7.19.24) (10.6.2,3) Rij = <e'i | R | uj> → <xei| R | tuj> // checked: <xei | R | tun> = Rin so OK <xei | R | tuj> = Rij So write (b) in all-Dirac notation: φ*T | xui> = Σj=1n <xei| R | tuj> |tuj> = Σj=1n Rij |tuj> We are trying to compare this to RT | xui> = Σj=1n | tuj><tuj| RT | xui> = Σj=1n <tuj | RT | xui> | tuj> = Σj=1n <xui| R |tuj > | tuj> = Σj=1n gij | tuj> = | tui> where I compute <xui | R | tun> = gin <xui | R | tuj> = gij . Since RT = R-1, this result makes perfect sense. So here is what I have so far φ*T | xui> = Rij |tuj> = linear combination of basis vectors in t-space RT | xui> = | tui> [ improved results 3.19.16 ] These are NOT the same, case closed. So the hunch that maybe φ*T = RT is wrong. We also know that (xui) = R(tui) or |xui> = |R(tui)> = R |tui> and |tui> = R-1|xui> = RT |xui> then can write first line as φ*T | xui> = Rij RT |xuj> or Rφ*T | xui> = Rij |xuj> which does not seem to add much. Now transpose the two equations above < xui| φ* = Rij <tuj| < xui| R = < tui| Close this with a t-space vector v to get < xui| φ*| v> = Rij <tuj|v> = Rijvj = [Rv]i (*) < xui| R | v> = < tui|v> = vi What destination am I seeking here? Since < xui| is a special case of <αx|, and since this is the goal <αx | φ*| vn,vn...vn> = <αx | Rvn,Rvn...Rvn> . or <αx | φ*| v> = <αx | Rv> . so the goal is <xui | φ*| v> = <xui | Rv> . (**) Statement of my Problem at 1 PM 3.19.16: How do you get from (*) to (**) ? Plan A: I now have a sandbox Section 10_6 v4.doc in which I make a different assumption about what φ* does. I will try that here and see what happens: Plan A 2. Want to test the following claim: [ try this in all-Dirac notation sep doc] (a) φ*T = RT and φ* = R [assume] for the pullback matrix. I think this is true from two sources: I derive it in "is phi star symmetric v2" It comes out at the end in "Section 10_6 v2". Now consider my definition of φ*T as stated in Ch 10 v 3 (b) φ*T(xei) ≡ Σj=1n Rij (tuj) i = 1..m [ok] (10.6.9) I claim that in Picture F' we have (7.19.24) (10.6.2,3) Rij = <e'i | R | uj> → <xei| R | tuj> // checked: <xei | R | tun> = Rin so OK <xei | R | tuj> = Rij So write (b) in all-Dirac notation: φ*T | xei> = Σj=1n <xei| R | tuj> |tuj> = Σj=1n Rij |tuj> We are trying to compare this to RT | xei> = Σj=1n | tuj><tuj| RT | xei> = Σj=1n <tuj | RT | xei> | tuj> = Σj=1n <xei| R |tuj > | tuj> = Σj=1n Rij | tuj> where I compute <xei | R | tun> = Rin <xei | R | tuj> = Rij . [ Since RT = R-1, this result makes perfect sense. not relevant now ] So here is what I have so far φ*T | xei> = Σj=1n Rij |tuj> = linear combination of basis vectors in t-space RT | xei> = Σj=1n Rij | tuj> [ improved results 3.19.16 3 PM ] These ARE the same, case closed. So the hunch that maybe φ*T = RT is right!. So this looks much better. Remember that (xui) = R(tui) so R maps the t-space basis vectors into the tangent base vectors for TxM in x-space. We can invert that to get (tui) = R-1(xui) = RT(xui) so RT does just the reverse. So φ* is just RT and it is a tempest in a teapot. There is no strange mysterious mapping. I guess I will now try a complete rewrite of Ch 10_6 with this new approach and see if it flies. Take a break at 3 PM.