try to show the two tensorized forms are the same v4

Show two tensorized forms are same v 4 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 = B'n;j;j . (15.2.8) I can simplify the first line using what I now know so it reads, (B)'n = (B'j;j);n – g'-1/2ε'nab(g'-1/2g'bcε'cdeB'e;d);a = (B'j;j);n – g'-1/2ε'nabg'-1/2ε'cdeg'bc B'e;d;a !! = (B'j;j);n – g'-1ε'nab ε'cdeg'bc B'e;d;a = (B'j;j);n – g'-1ε'nab ε'bde B'e;d;a So our entire task is reduced to showing this fact B'n;j;j = B'j;j;n – g'-1ε'nab ε'bde B'e;d;a ? We would like to get all the B indices to the same level, so let's choose B*** so start with B'n;j;j = B'j;j;n – g'-1ε'nab ε'bde B'e;d;a ? which I choose to rewrite as B'n;a;a = B'e;e;n – g'-1ε'nab ε'bde B'e;d;a ? Now we have three different "tilts" to worry about on the B indices. Let's process the first one B'n;a;a = δneB'e;a;a = δneδad B'e;d;a Now process the second term B'e;e;n = g'edB'e;d;n = g'edg'naB'e;da The problem is then to prove that this is true δneδad B'e;d;a = g'ed g'na B'e;d;a – g'-1ε'nab ε'bde B'e;d;a ? or [ δneδad – g'ed g'na + g'-1ε'nab ε'bde ] B'e;d;a = 0 ? or [ δneδad – g'ed g'na + g'-1ε'bna ε'bde ] B'e;d;a = 0 ? Now consider identity (D,11,10) which says |g'|-1 ε'cabε'ca'b' = δaa'δbb' – δab'δba' I will ignore the abs value and clear that up later. So assume the identity says g'-1 ε'cabε'ca'b' = δaa'δbb' – δab'δba' or g'-1 ε'bnaε'bde = δdnδea – δdaδen Then our equation of interest is this [ δneδad – g'ed g'na + δdnδea – δdaδen ] B'e;d;a = 0 or [ – g'ed g'na + δdnδea] B'e;d;a = 0 n fixed The bracket [..] does not vanish, but now consider [ – g'ed g'na + δdnδea] ReERdDRaA BE;D;A = 0 ? [ – g'ed g'na + δdnδea] ReERdDRaA BE,D,A = 0 ? [ – g'ed g'na + δdnδea] ReERdDRaA ∂D∂ABE = 0 ? – g'ed g'naReERdDRaA ∂D∂ABE +δdnδea ReERdDRaA ∂D∂ABE = 0 ? – RdERdDRnA ∂D∂ABE + RdERnDRdA ∂D∂ABE = 0 ? In the second term do D↔A and then reorder the derivatives – RdERdDRnA ∂D∂ABE + RdERnARdD ∂D∂ABE = 0 ? – (RdERdD)RnA ∂D∂ABE + (RdERdD) RnA ∂D∂ABE = 0 ? – (δED)RnA ∂D∂ABE + (δED) RnA ∂D∂ABE = 0 ? Since this last equation is obviously true, we go back up erasing question marks and we have then shown that the two tensorizations are in fact the same!