try to show the two tensorized forms are the same v2

Show two tensorized forms are same PhL 1.11.15 (B)'n = B'n;j;j = δijB'n;j;i = δij g'nmg'jk B'm;k;i = g'nm g'jk B'm;k;j (B)'n = (B'j;j);n – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a (I.5.1) // could say ,a at end Right at this point, we can simplify one thing, namely (g'-1/2g'bcε'cdeB'e;d);a = (g'-1/2ε'cdeB'e;d);a g'bc + g'bc;a (g'-1/2g'bcε'cdeB'e;d) = (g'-1/2ε'cdeB'e;d);a g'bc since we know that g'bc;a = 0. Note added 5/1/15. But now I know more, so further simplification is possible: = g'-1/2ε'cde (B'e;d);a g'bc Then our second form above is this (B)'n = (B'j;j);n – g'-1/2ε'nabg'bcg'-1/2ε'cde (B'e;d);a = δijB'j;i;n – g'-1/2ε'nabg'bcg'-1/2ε'cde (B'e;d);a = δij g'jm g'nsB'm;i;s – g'-1/2g'bcε'nabg'-1/2ε'cde(B'e;d);a = g'jm g'nsB'm;j;s – g'-1/2ε'nabg'bcg'-1/2ε'cde (B'e;d);a So show equal, I have to show that g'nm g'jk B'm;k;j = g'jm g'nsB'm;j;s – g'-1/2ε'nabg'bcg'-1/2ε'cde B'e;d;a where at least all the B' indices are in the same location (down). Rewrite g'nm g'jk B'm;k;j = g'km g'nsB'm;k;s – g'-1/2ε'nabg'bcg'-1/2ε'cde(B'e;d);a g'nm g'jk B'm;k;j = g'km g'njB'm;k;j – g'-1/2ε'nabg'bcg'-1/2ε'cde(B'e;d);a (g'nm g'jk - g'km g'nj )B'm;k;j = – g'-1/2ε'nabg'bcg'-1/2ε'cde(B'e;d);a (*) Now rewrite the RHS as – g'-1/2ε'nabg'bcg'-1/2ε'cde(B'e;d);a = - g'-1 ε'nabg'bc ε'cdeB'e;d;a = - g'-1 ε'njbg'bc ε'ckmB'm;k;j So now I have to show that (g'nm g'jk - g'km g'nj )B'm;k;j = - g'-1 ε'njbg'bc ε'ckmB'm;k;j or (g'nm g'jk - g'km g'nj + g'-1 ε'njbg'bc ε'ckm)B'm;k;j = 0 The proof would be all done if I can show that (g'nm g'jk - g'km g'nj + g'-1 ε'njbg'bc ε'ckm or g'nm g'jk - g'km g'nj + g'-1 ε'njc ε'ckm = 0 ? But I think I have proven this somewhere! Lower indices g'nm g'jk - g'km g'nj + g'-1 ε'njc ε'ckm = 0 ? Lower m and k g'nm g'jk - g'km g'nj + g'-1 ε'njc ε'ckm = 0 ? g'nm g'jk - g'km g'nj + g'-1 ε'njc ε'ckm = 0 ? It is close but I think something is wrong. Now the object on the right of (*) can be written g'bc(g'-1/2ε'cdeB'e;d);a = g'bc(g'-1/2ε'cde);aB'e;d + g'bc(g'-1/2ε'cde)B'e;d;a g'bc(g'-1/2ε'cdeB'e;d);a =g'bc (g'-1/2ε'cde);aB'e;d + g'bc(g'-1/2ε'ckm)B'm;k;a Then we have (g'nm g'jk - g'km g'nj )B'm;k;j = – g'-1/2ε'nab * {g'bc(g'-1/2ε'cde);aB'e;d +g'bc (g'-1/2ε'ckm)B'm;k;a or (g'nm g'jk - g'km g'nj )B'm;k;j = – g'-1/2ε'njb * {g'bc(g'-1/2ε'cde);jB'e;d + g'bc(g'-1/2ε'ckm)B'm;k;j } or (g'nm g'jk - g'km g'nj )B'm;k;j = – g'-1/2ε'njbg'bc(g'-1/2ε'cde);jB'e;d – g'-1/2ε'njb g'bc(g'-1/2ε'ckm)B'm;k;j Then we put all the B'm;k;j on the left [g'nm g'jk - g'km g'nj + g'-1/2ε'njb (g'-1/2g'bcε'ckm) ]B'm;k;j = – g'-1/2ε'njbg'bc(g'-1/2ε'cde);j B'e;d . or [g'nm g'jk - g'km g'nj + g'-1ε'cnj ε'ckm ] B'm;k;j (*) = – g'-1/2ε'njbg'bc(g'-1/2ε'cde);j B'e;d . It does not seem very likely! You would think the B'm;k;j would be independent of the B'e;d . No! that is not correct because B'm;k;j does contain some B'a;b terms in its Γ terms! Now the bracket on the left reminds me of this identity from Appendix D which is this (for g>0) g-1 εcabεca'b' = gaa'gbb' – gab'gba' (D.11.10) To make this match the existing ε'ε' structure appearing above do ab→nj and a'b' → km g'-1 ε'cnjε'ckm = g'nkg'jm – g'nmg'jk We can then rewrite (*) as [g'nm g'jk - g'km g'nj + g'nkg'jm – g'nmg'jk ] B'm;k;j (*) = – g'-1/2ε'njbg'bc(g'-1/2ε'cde);j B'e;d . or [ - g'km g'nj + g'nkg'jm ] B'm;k;j (*) = – g'-1/2ε'njbg'bc(g'-1/2ε'cde);j B'e;d . which at least simplifies things a bit. Now how about this (g'-1/2ε'cde);j = g'-1/2 (ε'cde);j + ε'cde (g'-1/2);j Now consider from theorem shown below (ε'cde);j = (1/) ∂'j() ε'cde So at this point let's install our known expansions: Ba;α = ∂α Ba – ΓnaαBn => B'e;d = ∂'d B'e – Γ'sedB's Ba;b;α = ∂α Ba;b – ΓnaαBn;b – ΓnbαBa;n => (F.9.20) B'm;k;j = ∂'j B'm;k – Γ'smjB's;k – Γ'skjB'm;s Installing these things, here is what we then have to show is true [g'nm g'jk - g'km g'nj + g'-1/2ε'njb (g'-1/2g'bcε'ckm)] [ ∂'j B'm;k – Γ'smjB's;k – Γ'skjB'm;s ] + g'-1/2ε'njb(g'-1/2g'bcε'cde);j [ ∂'d B'e – Γ'sedB's] = 0 which we can simplify a bit to read [g'nm g'jk - g'km g'nj + g'-1ε'njbg'bcε'ckm)] [ ∂'j B'm;k – Γ'smjB's;k – Γ'skjB'm;s ] + g'-1ε'njb(g'bcε'cde);j [ ∂'d B'e – Γ'sedB's] = 0 Now I call upon my Appendix D identity which says gaa' gbb' – gab' gba' - g-1 gdc εcabεda'b' = 0 which I translate to read gnm gjk – gnk gjm - g-1 gdc εcnjεdmk = 0 Verify this first for Cartesian just as a check: [δ'nm δ'jk - δ'km δ'nj + ε'njb (ε'bkm)] [ ∂'j B'm,k ] + (ε'bnjε'bde),j [ ∂d Be] = 0 ? The RHS vanishes since ε' = constant, and LHS probably OK for same reason shown earlier. Now back to (**). One simplification is this: (g'-1/2g'bcε'cde);j = (g'-1/2ε'cde);j g'bc + (g'-1/2ε'cde)g'bc;j = (g'-1/2ε'cde);j g'bc = ε'cde (g'-1/2);j g'bc = ε'cde (g'-1/2),j g'bc = ε'cde ∂'j(g'-1/2) g'bc So then we want to show this to be true: [g'nm g'jk - g'km g'nj + g'-1/2ε'njb (g'-1/2g'bcε'ckm)] [ ∂'j B'm;k – Γ'smjB's;k – Γ'skjB'm;s ] + g'-1/2ε'njb ε'cde ∂'j(g'-1/2) g'bc [ ∂'d B'e – Γ'sedB's] = 0 (**) Now remove all primes just to make things easier, [gnm gjk - gkm gnj + g-1/2εnjb (g-1/2gbcεckm)] [ ∂j Bm;k – ΓsmjBs;k – ΓskjBm;s ] + g-1/2εnjb εcde ∂j(g-1/2) gbc [ ∂d Be – ΓsedBs] = 0 ? Now there is no alternative but to expand all three Bm;k terms like this: Ba;α = ∂α Ba – ΓraαBr => Bm;k = ∂k Bm – ΓrmkBr Bs;k = ∂k Bs – ΓrskBr Bm;s = ∂s Bm – ΓrmsBr Then we have to show that [gnm gjk - gkm gnj + g-1/2εnjb (g-1/2gbcεckm)] * { ∂j * [ ∂k Bm – ΓrmkBr] - Γsmj[ ∂k Bs – ΓrskBr] - Γ'skj [∂s Bm – ΓrmsBr] } + g-1/2εnjb εcde ∂j(g-1/2) gbc [ ∂d Be – ΓsedBs] = 0 If this is true, then we must have the ∂j∂kBm term vanish separately! That term is [gnm gjk - gkm gnj + g-1/2εnjb (g-1/2gbcεckm)] ∂j∂kBm = 0 ? [gnm gjk - gkm gnj + g-1εnjb gbcεckm] ∂j∂kBm = 0 ? How does this work in Cartesian> [δnm δjk - δkm δnj + εbnj (εbkm)] ∂j∂kBm = 0 ? and I can see how that goes. In the general case I guess we must show that [] is antisym, so [gnm gjk - gkm gnj + g-1εnjb gbcεckm] = - [gnm gjk - gjm gnk + g-1εnkb gbcεcjm] ? [gnm gjk - gkm gnj + g-1εnjb gbcεckm] = - gnm gjk + gjm gnk - g-1 εnkb gbcεcjm ? gnm gjk - gkm gnj + gnm gjk - gjm gnk = - g-1gbc [εnjbεckm + εnkbεcjm] ? 2 gnm gjk - gkm gnj - gjm gnk = - g-1gbc [εbnjεckm + εbnkεcjm] ? That is a new one on me! Why should this be true? Go back to εbacεbde = δadδce - δaeδcd = δadδce - δaeδcd εbacεbde = = gadgce - gaegcd Now add g-1on the right to get weights the same I may already have the answer to this mystery in Appendix D!! My example at (D.11.2) starts with εabεa'b' = = δa,a' δb,b' – δa,b' δb,a' But this could just as well be εcabεca'b' = = δa,a' δb,b' – δa,b' δb,a' Write this in Cartesian space as g-1 εcabεca'b' = gaa' gbb' – gab' gba' Weights match so we are tensorized. Now raise indices g-1 εcabεca'b' = gaa' gbb' – gab' gba' or gaa' gbb' – gab' gba' = g-1 gdc εcabεda'b' So there you are! Appendix 1. What can I say about (ε'abc);d ? On the one hand you want to argue that ε'abc is a constant like π, and you know (π);d = π,d = 0 But on the other hand ε' is a tensor, so we need to do the full deal. Start with Babc;α ≡ ∂α Babc – ΓnaαBnbc – ΓnbαBanc – ΓncαBabn (F.9.10) Now raise indices using the rules (F.8.9), Babc;α ≡ ∂α Babc + ΓanαBnbc + ΓbnαBanc + ΓcnαBabn (F.9.10) Then we would have εabc;α ≡ ∂α εabc + Γanαεnbc + Γbnαεanc + Γcnαεabn (F.9.10) = Γanαεnbc + Γbnαεanc + Γcnαεabn = Γanαεnbc + Γbnαεnca + Γcnαεnab and this does NOT seem to vanish! Both sides are antisymmetric in a,b,c. But I often say there is only one such tensor, so we must have εabc;α = Γanαεnbc + Γbnαεnca + Γcnαεnab = Cαεabc // suspected step!! Now try to evaluate Cα: Take abc = 123 ε123;α = Γ1nαεn23 + Γ2nαεn31 + Γ3nαεn12 = Cε123 or ε123;α = Γ11αε123 + Γ22αε231 + Γ33αε312 = C or ε123;α = Γ11α + Γ22α + Γ33α = Cα Therefore Cα = Γiiα But I know that Γaan = (1/2) gad ∂ngad = (1/2)(1/g)∂ng = (1/) ∂n() . (F.4.2) And therefore Cα = Γiiα = (1/) ∂α() and therefore εabc;α = Cαεabc = (1/) ∂α()εabc This might be worth adding somewhere. Appendix 2: What can we say about (g'-1/2);j ? This is has weight +1 so I think (g'-1/2);j = ? The weigh of the scalar density is W = +1 so B;α = ∂αB + W Γκκα B (F.9.23) Γκκα = (2g)-1∂αg from (F.4.2) says that (g'-1/2);j = ∂j(g'-1/2) + (1) (2g')-1(∂'jg') (g'-1/2) = (-1/2) g'-3/2 (∂jg') + (2g')-1(∂'jg') (g'-1/2) = (-1/2) g'-3/2 (∂jg') + (1/2) (g')-3/2(∂'jg') = 0 surprise!! There must be some easier way to know this fact! I do know this, gab;c = 0 weight g = -2 g = det(gab) = εabcg1ag2b What about this ;j = ∂'j() - Γκκα = ∂'j() - (1/) ∂j() = 0 g;j = ∂jg + W Γκκα g = ∂jg - 2 Γκκα g = ∂jg - 2(1/2)(1/g)∂jg g = ∂jg - (1/g)∂jg g = 0 Why is it that g;j = 0 ? Here is an idea det(Mij) = (1/g) (1/N!) εab..x εAB...X MAa MBb ...MXx (D.12.15) g = det(gij) = a different story! How modify (D.12.* results)? det(Mij) = εab..x M1aM2b....MNc (D.12.1) = εab..x M1aM2b....MNc How about this: det(Mij) = g det(Mij) // scalar density of weight -2 g = det(gij) (D.12.18) det(Mij) = (1/g) (1/N!) εab..x εAB...X MAa MBb ...MXx How apply this to the metric tensor? det(gij) = g det(gij) // scalar density of weight -2 g = det(gij) (D.12.18) det(gij) = (1/g) (1/N!) εab..x εAB...X gAa gBb ...gXx = (1/g) (1/N!) εab..x εAB...X δAa δBb ...δXx = (1/g) (1/N!) εab..x εab...x = (1/g) Then I get det(gij) = g det(gij) = g (1/g) = 1 = wrong!!!!