try to show the two tensorized forms are the same v3

Show two tensorized forms are same v 3 PhL 1.11.15 Here are the "two forms" for (B)'n (B)'n = (B'j;j);n – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a (15.7.9) (B)n = Bn;j;j . (15.2.8) We want to show these are the same. Remove all primes and we want to show that (B)n = (Bj;j);n – g-1/2εnab(g-1/2gbcεcdeBe;d);a (15.7.19) (B)n = Bn;j;j . (15.2.8) I can simplify the first line using what I now know so it reads, (B)n = (Bj;j);n – g-1/2εnab(g-1/2gbcεcdeBe;d);a = (Bj;j);n – g-1/2εnabg-1/2εcdegbc Be;d;a !! = (Bj;j);n – g-1εnab εcdegbc Be;d;a = (Bj;j);n – g-1εnab εbde Be;d;a So our entire task is reduced to showing this fact Bn;j;j = Bj;j;n – g-1εnab εbde Be;d;a ? We would like to get all the B indices to the same level, so let's choose B*** so start with Bn;j;j = Bj;j;n – g-1εnab εbde Be;d;a ? which I choose to rewrite as Bn;a;a = Be;e;n – g-1εnab εbde Be;d;a ? Now we have three different "tilts" to worry about on the B indices. Let's process the first one Bn;a;a = δneBe;a;a = δneδad Be;d;a Now process the second term Be;e;n = gedBe;d;n = gedgnaBe;da The problem is then to prove that this is true δneδad Be;d;a = ged gna Be;d;a – g-1εnab εbde Be;d;a ? or [ δneδad – ged gna + g-1εnab εbde ] Be;d;a = 0 ? or [ gnegad – ged gna + g-1εnab εbde ] Be;d;a = 0 ? This will be proved if we can show that gnegad – ged gna + g-1εnab εbde = 0 ? or gnegad – ged gna + g-1εbna εbde = 0 ? εnab = εbna I don't think this is true, so we have a problem. At least it is a tensor equation in terms of index positions. Here is something I think IS true from above (D.11.10) |g|-1 εcabεca'b' = δaa'δbb' – δab'δba' = gaa'gbb' – gab'gba' Ignoring the abs value for now, this says g-1 εcabεca'b' = gaa'gbb' – gab'gba' or g-1 εbnaεbde = gdngea – gdagen I think this is a true equation. Then I have to show this gnegad – ged gna + gdngea – gdagen = 0 or – ged gna + gdngea = 0 or ged gna = gdngea or ged gna = δdnδea But I just don't see how this can be true!! Is this some Fact that I have missed? I do know g** = (g**)-1 = cof(g**)/det(g**) = cof(g**)/ g So this says gna = cof(gna)/ g and then ged gna = ged cof(gna)/ g Suppose the coordinate system is orthogonal! Then consider ged gna = he2δed hn-2δna = (he2hn-2) δedδna but this is a far cry from what I need to be true. Back up and ponder uniqueness of tensorization Consider these two paths: (∂j2)Bn = ∂j∂j Bn = Bn,j,j → Bn;j;j (∂j2)Bn = ∂j∂j Bn = Bn,j,j → Bn;j;j My claim that "tensorization is unique" would then imply that Bn;j;j = Bn;j;j which seems reasonable from my reverse-tilt-rule contraction theorem. But consider Bn;j;j = Bn;a;a = gab Bn;b;a = gabgac Bn;b;c = δbc Bn;b;c = Bn;b;b = Bn;jb;j QED So what is my problem? Try a different case: ∂n(∂jBj) = (Bj,j),n → (Bj;j);n ∂j(∂nBj) = Bj,n,j → Bj;n;j My claim that "tensorization is unique" would then imply that (Bj;j);n = Bj;n;j ? (*) This is no longer a simple case of the reverse tilt rule. And this claim is NOT TRUE as I will now show. First, process the left side of (*) (Bj;j);n = Σj(Bj;j);n = Σj(ΣiδjiBi;j);n = Σij δji(Bi;j;n) = Σij δji(Σm gjmBi;m;n) = Σijmδjigjm (Bi;m;n) = Σijmδjigjm (ΣsgnsBi;ms) = Σijms δjigjmgns (Bi;ms) = Σjms gjmgns (Bj;ms) Now process the right side of (*) ΣjBj;n;j = Σjmsδnm δjs Bj;m;s then (*) becomes Σjms gjmgns (Bj;ms) = Σjmsδnm δjs Bj;m;s ? or Σjms[ gjmgns - δnm δjs] Bj;m;s = 0? For this to be true for every tensor B, we would need to have gjmgns = δnm δjs But this is NOT TRUE. Conclusion: My claim of unique tensorization stated near (15.2.5) is wrong! Something then is wrong with my proof. Let's then step through my proof: {...}1a = {...}2a when both sides are evaluated in Cartesian x-space . (15.2.5) (Bj;j);a = Bj;a;j // realization of the above! "One can then transform both objects to x'-space in the usual manner, {...}'1a = Rab{...}1b and {...}'2a = Rab{...}2b . " (15.2.6) (B'j;j);a = Rab(Bj;j);b and B'j;a;j = Rab Bj;b;j "Therefore, since {...}1a = {...}2a in x-space, one must have {...}'1a = {...}'2a in x'-space. " Well, I would write (B'j;j);a = Rab(Bj,j),b B'j;a;j = Rab Bj,b,j (*) And it is true the that right sides are indeed the same since in Cartesian space (Bj,j),b = ∂b(∂jBj) Bj,b,j = ∂j(∂bBj) Since the right sides of (*) are exactly the same in Cartesian space, the left sides must also be equal and we get (B'j;j);a = B'j;a;j So I see the claimed logic of my proof!