paradox v1 3_14_16 REVIEWED

Paradox v1 3_14_16 involving Section 6.10 PhL 3.14.16 5.17.16: I did have this horrible paradox starting on page 4 below. It turned out that it was a confusion about the meaning of a component label, and that subject was cloudy at the time, now it is all axis-aligned default meaning. I have other doc(s) where this same paradox is addressed. At the same time, below I am still thinking of φ* as an operator instead of a function. Also, early covariant transpose comments right below. [3.17.16 I am going through this doc with a fine tooth comb now that I understand the paradox problem.] 1. Theorem: RTR = 1 even when R is non-square (covariant transpose T) Proof. We know that Rab ≡ (∂xa/∂tb) for full range of all indices. The question with covariant T is then this: RTR = 1 ? (RT)abRbc = δac ? Rba Rbc = δac ? // now use Tensor (7.5.16) Σb(∂xa/∂x'b) (∂x'b/∂xc) = δac true, this is the chain rule even for non-square Alternative proof: RTR = 1 ? SR = 1 ? true 2. Plan A. Now consider my definition of φ*T as stated in Ch 10 Pause: What are the "natural bases" in this context? I now comment on this in (10.6.4) and the answer is that the two natural bases are tua for t-space and xea for x-space (just new names for things). (b) φ*T(xui) ≡ Σj=1n Rij (tuj) i = 1..m (**) (10.6.9) Dot this with (tua) to get < tua | φ*T | xui> = Ria [ok 3.17] or < xui | φ* | tua> = Ria . i = 1..m, a=1..n (1) We know that (xui) = R(tui) but only for i = 1..n. The (xui) for i=n+1..m are GSO determined. So for i = 1..n we can claim from (**) above [φ*TR](tui) = Σj=1n Rij (tuj) i = 1..n [ok 3.17] Dot with (tur) to get (tur) [φ*TR](tui) = Σj=1n Rij (tur) (tuj) = Rir i,r = 1..n or < tur | φ*TR | tui> = Rir i,r = 1..n (2) [ok 3.17] Meanwhile, what is this matrix element: ? < tur | R | tui> = ? nothing, conformance mismatch! What about this similar matrix element: < xur | R | tui> = < xur | R tui> = < xur |xui> = δri [ok 3.17] Can rewrite the first above flipping the scalar product, < tui | RTφ* | tur> = Rir i,r = 1..n (3) [ok 3.17] and here the tilt looks better. Just for fun you could throw in 1 = Σa=1m |xua><xua| to get Σa=1m < tui | RT|xua><xua| φ* | tur> = Rir ? i,r = 1..n Σa=1m < Rtui |xua><xua| φ* | tur> = Rir ? i,r = 1..n Σa=1m < Rtui |xua> Rar = Rir ? i,r = 1..n Σa=1m < xui |xua> Rar = Rir ? i,r = 1..n Σa=1m δia Rar = Rir ? i,r = 1..n Rir = Rir ? i,r = 1..n // yes!!! so consistent. Summary to this point: < xui | φ* | tua> = Ria . i = 1..m, a=1..n (1) [ok 3.17] < tui | RTφ* | tur> = Rir i,r = 1..n (3) [ok 3.17] <xui | R | tur> = <xui | xur> = δir i = 1,2..m r = 1,2..n [ok 3.17] where last is derived in line. So at least we have three interesting matrix elements. I'm not sure what to do with any of them right now. But I have gotten rid of any notion that φ* = R as operators. Go back to a few inches above, < xui | φ* | tua> = Ria . i = 1..m, a=1..n (1) Rewrite the RHS as { note that index "a" in [Rtui]a is in the natural basis which is the xea basis } Ria = [Rtui]a // = Rak (tui)k Then we have this version of (1) < xui | φ* | tua> = [R tui]a [ok 3.17 with meaning of a noted above ] Now write |v> = Σj vj tuj [ok 3.17, a vector in t-space] Then < xui | φ* |v> = Σj vj < xui | φ* | tuj> = Σj vj Rij = ΣjRijvj = [Rv]i [natural] (***) [ok 3.17 understanding that "i" is the natural index which is for the xea natural basis in x-space ] But < xui | R |v> = < xui | Rv> = [Rv]i [un-natural] (****) [ok 3.17 understanding that "i" is the UN- natural index which is for the xui natural basis in x-space ] [ thus the two objects above both called [Rv]i are not the same object ! ] So we are finding that < xui | φ* |v> = < xui | R |v> = [Rv]i [ This is WRONG because the two index i's are different as noted above! ] So somehow in this narrow situation, we can in fact replace φ* by R and get the same result.[ wrong!] This is I think the correct result and no doubt agrees with the full result at the end of "Section 10_6 v3" doc which says <αx | φ*| vn,vn...vn> = <αx | Rvn,Rvn...Rvn> . [ Idea: Maybe I should have xei on the left instead of xui in an αx expansion? ] Go back to simpler case < xui | φ* | tua> = Ria . i = 1..m, a=1..n (1) [ok 3.17] < xui | R | tua> = < xui | xua> = δia does not work here [ok 3.17] How can it work for a general vector v but not work for a basis vector tua ??? [ It doesn't work for the general vector v, as noted above. ] Take the v result and replace v by tua and first translate (***) < xui | φ* |v> = Σj vj < xui | φ* | tuj> = Σj vj Rij = ΣjRijvj = [Rv]i [natural] (***) < xui | φ* |tua> = Σj δaj < xui | φ* | tuj> = Σj δaj Rij = ΣjRijδaj = Ria Then translate (****) < xui | R |v> = < xui | Rv> = [Rv]i [unnatural] (****) < xui | R | tua> = < xui | R tua> = [R tua]i = Rij(tua)j = Ria // same result as (***) [unnatural] so [wrong] [wrong] But this conflicts with < xui | R | tua> = < xui | xua> = δia [right] so we are back to having a Paradox! Resume here manana. // Manana is here, I resume: [ 3.17 Since the previous equation is wrong, there is no paradox. ] This paradox is very isolated! Let's review the last two lines step by step. Here is the first line in steps: Paradox of 3/15/16 Pretty simple to present, and assume simple case that m = n. On the one hand, according to Argument 1: < xui | R tua> // vector on the right exists in x-space 1 = [R tua]i // idea that < xui | q > = qi for any vector in x-space (in xui expansion) [unnatural] 2 = Σj=1n Rij(tua)j // matrix * vector multiplication [so wrong here] 3 = Σj=1n Rij δaj // definition of the basis vectors in t-space 4 = Ria // use up the delta function On the other hand, according to Argument 2 Trying to concentrate the paradox more < xui | R tua> // vector on the right exists in x-space 5 = < xui | xua> // idea that R |tua> = | xua> for a = 1..n, definition of the | xua> // this is (10.6.4) of Section 10_6 v.3.doc, but up instead of down 6 = δia [right] // property of regular with dual vectors (see Task A below) There you have it!!! We get conflicting evaluations of this matrix element of R ! I am now searching for what might be wrong, taking a break here at 4:10 PM 3.17 // Resume at 6:20: Proposed Task A. Show the that xuj and xui really satisfy the dual property <xub | xua> = δba given the following two presumed definitions (presumed to be compatible in this dual sense) (xua) ≡ R(tua) (xub) ≡ R(tub) a,b = 1..n or | xua> = R | tua> | xub> = R | tub> Write out each equation from the first line above. Here is what I think each of those equations says: (xua)i ≡ Σj=1n Rij(tua)j (xub)i ≡ Σj=1n Rij(tub)j a,b = 1..n = Ria = Rib Now we test the rule and here T refers to the covariant transpose, <xub | xua> = Σi=1m (xub)i (xua)i = Σi=1m Rib Ria = Σi=1m (RT)bi Ria = (RTR)ba = δba so Task A is concluded with the desired result, where a and b are restricted to the range 1..n. Looking at Arg 1 and Arg 2, it would seem that something is wrong in one of the steps 1,2,3,4 of Arg 1 [correct], or alternatively, the idea that RTR = 1 is not valid for n ≠ m [incorrect]. But we can always take m = n as a test case and we still have this problem! So this paradox exists within the regular square matrix world too! [correct] Examine Step 1 of Arg 1: = < xui | R tua> // vector on the right exists in x-space, allowed to move R as shown 1 = [R tua]i // idea that <xui|xq> = xqi for any vector in x-space (in xui expansion) The idea is that we expand any vector xq which lies in the tangent space subspace in this manner: xq = Σi=1n xqi (xui) [ = Σi=1n [xqi](xu) (xui) unnatural ] [ (xqi) = xq (xui) ] Only the first n basis vectors in x-space contribute to such a vector. Now such a vector would be xq = [R (tua)] [ok 3.17, a sample vector in x-space] |xq> = R |tua> = | xei>< xei| R |tua> = Ria | xei> [ok] [ok] [ok] This vector certainly is an element of TxM since in fact it is just (xua) which in fact is one of the tangent space basis vectors. So let's apply the above expansion to this vector xq = Σj=1n xqj (xuj) [R (tua)] = Σj=1n [R (tua)](xu)j (xuj) // = (xua) we know [ok] Using the dual property of the xui and xuj vectors as just verified in Task A above, we get (xui) [R (tua)] = (xui) { Σj=1n [R(xu) (tua)]j (xuj)} = Σj=1n [R(xu) (tua)]j δij = [R(xu) (tua)]i or < xui | R tua> = [R(xu) (tua)]i and so we have verified Step 1. Examine Step 2 of Arg 1: 1 = [R(xu) tua]i // idea that < xui | q > = qi for any vector in x-space (in xui expansion) 2 = Σj=1n Rij(tua)j // matrix * vector multiplication [ wrong, R not in natural basis ] Well, R is a matrix and tua is a vector with n components, and this is how we multiply a matrix by a vector on the right. How do I know that the tilt is as shown? When we write Vx = R Vt in Picture F, we always write (Vx)i = Rij (Vt)j and (Vx)i = Rij (Vt)j for contra and co vector transformations. The above is the second case so the tilt is correct. Examine Steps 3 and 4 of Arg 1: [ these steps are all OK, it is step 2 above that is wrong ] 2 = Σj=1n Rij(tua)j // matrix * vector multiplication 3 = Σj=1n Rij δaj // definition of the basis vectors in t-space 4 = Ria // use up the delta function Well, by definition the t-space basis vectors have the property (tua)j = δaj. This goes back to (2.4.1) of wedge doc (un)i = δni (un)i = δni (7.18.3)' (2.4.1) So there is nothing wrong with steps 3 and 4. Let's rephrase this paradox in regular Picture A context of tensor doc with m = n Tensor Doc Paradox of 3/15/16 Pretty simple to present, and we assume simple case that m = n. In this case, we have this picture and situation: x' = F(x) Rab ≡ (∂x'a/∂xb) = ∂bx'a Sab ≡ (∂xa/∂x'b) = ∂'bxa e'a with (e'a)b = δab axis-aligned basis vectors in x'-space ea ea= Se'a tangent base vector in x-space ua with (ua)b = δab axis-aligned basis vectors in x-space u'a u'a= Rua tangent base vector in x'-space (3.5.3) (10.6.2) Here then is the paradox in this notation On the one hand, according to Argument 1: < u'i | R ua> // vector on the right exists in x'-space 1 = [R ua](u')i // idea that < u'i | q > = qi for any vector in x'-space (in u'i expansion) 2 = Σj=1n Rij(ua)j // matrix * vector multiplication [ wrong because e' is the natural basis] 3 = Σj=1n Rij δaj // definition of the basis vectors in x-space 4 = Ria // use up the delta function On the other hand, according to Argument 2 Trying to concentrate the paradox more < u'i | R ua> // vector on the right exists in x-space 5 = < u'i | u'a> // idea that R |ua> = | u'a> for a = 1..n, definition of the | u'a> // this is (10.6.4) of Section 10_6 v.3.doc, but up instead of down 6 = δia // property of regular with dual vectors (see Task A below) There you have it!!! We get conflicting evaluations of this matrix element of R ! Let's rewrite this paradox in the other direction which should be most familiar to me. Do these rules, g'↔ g R ↔ S en → u'n e'n → un en → u'n e'n → un (7.18.2) but backwards. Then get [ OK don't bother with this one ] On the one hand, according to Argument 1: < ei | Se'a> // vector on the right exists in x-space 1 = [S e'a](e)i // idea that < ei | q > = qi for any vector in x'-space (in ei expansion) 2 = Σj=1n Sij(e'a)j // matrix * vector multiplication 3 = Σj=1n Sij δaj // definition of the basis vectors in x'-space 4 = Sia // use up the delta function On the other hand, according to Argument 2 Trying to concentrate the paradox more < ei | S e'a> // vector on the right exists in x-space 5 = < ei | ea > // idea that S |e'a> = | ea> for a = 1..n, definition of the |ea> // this is (10.6.4) of Section 10_6 v.3.doc, but up instead of down 6 = δia // property of regular with dual vectors (see Task A below) This form does bring up some issues I can now consider. At the start < ei | Se'a> // vector on the right exists in x-space 1 = (S e'a)(e)i // idea that < ei | q > = qi for any vector in x'-space (in ei expansion) This says the this is a component of type V(e)i and not V(u)i as in wedge doc Chapter 2. The relevant expansion is this, S e'a = Σi (S e'a)(e)i ei = Σi (S e'a)(u)i ui Now what can we say about these two objects, (S e'a)(e)i (S e'a)(u)i How do you write out the "matrix multiplication" here? I think you would say (S e'a)(e)i = Σj Sij (e'a)(e)j but how do I know that is correct? Many issues are appearing now that amazingly seem hazy after all the years I have spent doing this stuff. When I write the very normal w = Mv it is an abbreviation for the set of equations, wi = Mijvj where wi means <ui|w> with Cartesian aligned axes basis vectors ui and Mij are just some numbers. If you rotate the axes [ Resume now below which is older than the above and before I had a paradox. ] _______________________________________ 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 (b) φ*T(xui) ≡ Σj=1n Rij (tuj) i = 1..m [ok] (10.6.9) Based on what is stated above (that φ*T = RT), this implies that (c) RT | xui> = Σj=1n Rij | tuj> . [ok] In regular matrix notation this would say RT (xui) = Σj=1n Rij (tuj) [ok] Writing out components on both sides, [RT (xui)](tu)s = Σj=1n Rij (tuj)(tu)s [ assume s is "natural" ] Pause. Note that the natural Rij is this: Rij = <e'i | R | uj> → <xei| R | tuj> = [R(xe,tu)]ij so transpose the matrix element to get Rij = <tuj| RT | xei> = [RT(tu,xe) ]ij So I would say [RT (xui)](tu)s = <tujs | RT | xui> = < xui | R |tujs> = [R(xu,tu)]is = unnatural ≠ Ris Add detail to the left (RT)sr (xui)r = Σj=1n Rij (tuj)s [RT (xui)](tu)s = [RT(tu,xu)]sr (xui)(xu)r But we have already shown that (xui)r = Rri and (tuj)s = δjs so the above says (RT)sr Rri = Ris which says (RTR)si = Ris or δsi = Ris which is WRONG. It is wrong if I pursue various other paths as well. Can I prove this claim directly? I know from (10.6.6) that (d) (xui) = R(tui) or |xui> = R| tui> i = 1..n Inserting this for |xui> in the previous result gives RT [ R| tui>] = Σj=1n Rij | tuj> or (e) RTR | tui> = Σj=1n Rij | tuj>. STOP RIGHT HERE. Since RTR = 1 this says | tui> = Σj=1n Rij | tuj> which is wrong. No need to continue. Now close this with basis vector <tus| to get (f) <tus| RTR| tui> = Σj=1n Rij <tus| tui> = Ris since <tus| tui> = δsi. Now move RT to the left as usual to get (g) <Rt us | R tui> = Σj=1n Rij <tus| tui> = Ris which is just a rewrite. Now, as noted above, |xei> = R| tui> = | R tui> <xei| = < R tui | (h) <xes| = < R tus | Then (i) <Rt us | R tui> = <xes| xei> = δsi 3. So the paradox is that I have these two conflicting results. <Rt us | R tui> = Ris <Rt us | R tui> = δsi One possible resolution would be this φ*T = RT is WRONG. Comment 1. Recall from 10.6 that [φ*T(xei)]s = Ris . (10.6.11) How would you write the LHS? < tus| [ φ*T | xei>] = Ris ? < tus | φ*T | xei>] = Ris ? <xei | φ* | tus>] = Ris ? real scalar product Plan A: If you define ei the right way, it MUST do ei ej = δij. So now did I define ei in this case? Go back to this statement from 10_6 v2: which defines xei . (xei) = R(tui) or (xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.4) How might we raise the label i ? From the second line it would seem that (xei)j = Rji (xei)j = Rji and (xei)j = Rji but the first line is less obvious. Maybe rewrite the second line as (xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji Then you could write (xei) = Rut (tui) and then we have these two lines: (xei) = Rdt(tui) (xei) = Rut(tui) Summarizing the second case: (xei) = Rut(tui) (xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji and this agrees with what I had. Plan A: If you define ei the right way, it MUST do ei ej = δij. So now did I define ei in this case? I start with this: (xei) = R(tui) as definition of the xei . I write this out as (xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.4) How might we raise the label i ? From the second line it would seem that (xei)j = Rji (xei)j = Rji and (xei)j = Rji but the first line is less obvious. Maybe rewrite the second line as (xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji Then you could write (xei) = Rut (tui) and then we have these two lines: (xei) = Rdt(tui) (xei) = Rut(tui) Summarizing the second case: (xei) = Rut(tui) (xei)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji and this agrees with what I had.