Sept 11, 2008

_{Quotations from Evans' blog in} black

4) Finally Bruhn attempts to prove the Bianchi identity:

                D ^ T^a := R^a_b ^ q^b                                                 (7)

He ignores the detailed proofs I have given, and asserts notwithstanding that I have no knowledge of differential geometry. This is despite the fact that my proofs have been well accepted worldwide. The proof given by Bruhn of eq.(7) is as follows:

                D ^ T^a = d ^ T^a + ω^a_b ^ T^b
                                = d ^ (d ^ q^a + ω^a_b ^ q^b) + ω^a_b ^ (d ^ q^b + ω^b_c ^ q^c)

It is correct to this point

Thanks, Myron!!!

but then he jumps to

                D ^ T^a =? d ^ ω^a_b + ω^a_b ^ (ω^b_c ^ q^b) + ω^a_b ^ (d ^ q^b).

Then there is another jump to eq.(7).

Excuse me, but I did ignore that you have discarded the Poincaré rule in your New Math:

                d Ù (d Ù q^a) = 0 ,

to obtain from the last line accepted by you

                D Ù T^a = 0 + d Ù (ω^a_b Ù q^b) + ω^a_b Ù (d Ù q^b + ω^b_c Ù q^c) .

The last step is an obvious application of the Leibniz rule as valid in conventional math:

                d Ù (ω^a_b Ù q^b) = (d Ù ω^a_b) Ù q^b − ω^a_b Ù d Ù q^b ,

hence

                D Ù T^a = (d Ù ω^a_b) Ù q^b − ω^a_b Ù d Ù q^b + ω^a_b Ù (d Ù q^b + ω^b_c Ù q^c)
                                = (d Ù ω^a_b) Ù q^b + ω^a_b Ù ω^b_c Ù q^c
                                = (d Ù ω^a_b) Ù q^b + ω^a_c Ù ω^c_b Ù q^b
                                = (d Ù ω^a_b + ω^a_c Ù ω^c_b) Ù q^b = R^a_b Ù q^b .

by using the definition R^a_b := d Ù ω^a_b + ω^a_c Ù ω^c_b .

So on scholarly inspection this ''proof'' is not a proof at all. The correct way to prove eq.(7) is to expand it ... in tensor notation and use the tetrad postulate.

Can we agree, Myron, that the above complete proof is valid only in conventional math, while some results are unattainable in New Math (e.g. the 2nd Bianchi identity under general torsion)?