The first Bianchi identity in Conventional Math

a commentary on
Evans' webnote (p.4/no.4) on New Math proofs of the 1st Bianchi

Gerhard W. Bruhn, Darmstadt University of Technology

Sept 11, 2008

Quotations from Evans' blog in

4) Finally Bruhn attempts to prove the Bianchi identity:

                D ^ Ta := Rab ^ qb                                                 (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 ^ Ta = d ^ Ta + ωab ^ Tb
                                = d ^ (d ^ qa + ωab ^ qb) + ωab ^ (d ^ qb + ωbc ^ qc)

It is correct to this point

Thanks, Myron!!!

but then he jumps to

                D ^ Ta =? d ^ ωab + ωab ^ (ωbc ^ qb) + ωab ^ (d ^ qb).

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 qa) = 0 ,

to obtain from the last line accepted by you

                D Ù Ta = 0 + d Ùab Ù qb) + ωab Ù (d Ù qb + ωbc Ù qc) .

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

                d Ùab Ù qb) = (d Ù ωab) Ù qb − ωab Ù d Ù qb ,


                D Ù Ta = (d Ù ωab) Ù qb − ωab Ù d Ù qb + ωab Ù (d Ù qb + ωbc Ù qc)
                                = (d Ù ωab) Ù qb + ωab Ù ωbc Ù qc
                                = (d Ù ωab) Ù qb + ωac Ù ωcb Ù qb
                                = (d Ù ωab + ωac Ù ωcb) Ù qb = Rab Ù qb .

by using the definition Rab := d Ù ωab + ωac Ù ωcb .

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)?

Appendix: Cartan differential geometry using non-coordinate bases

Having understood now the problems of a New Math follower with Cartan's calculus of differential forms let me sketch Cartan's method of non-coordinate bases in differential geometry (see also the Appendix J of [1]).

Given on a n-dimensional manifold M a non-coordinate basis of 1-forms

                qa := qaα dxα                         (a,α=1,...,n)                                                 (A1)

which, in a certain sense, replaces the basis of coordinate 1-forms dxα (α=1,...,n).

Given in addition a connection by a set of n˛ 1-forms

                ωab := ωαab dxα                 (a,b,α=1,...,n)                                                 (A2)

The Greek indices refer to the basis of the coordinate 1-forms dxα, the Latin indices refer to the basis of the non-coordinate 1-forms qa.

Now introduce two sets of 2-forms:

1) n torsion forms Ta by

                Ta := d Ù qa + ωab Ù qb                         (a=1,...,n).                                 (A3)

2) n˛ curvature forms Rab by

                Rab := d Ù ωab + ωac Ù ωcb                 (a,b=1,...,n).                                 (A4)

This is the simple basis of a non-coordinate differential geometry on the manifold M.

As shown above a 1st Bianchi identity can be derived:

                D Ù Ta := d Ù Ta + ωab Ù Tb = Rab Ù qb .                                                 (A5)

And in addition a 2nd Bianchi identity as shown in Proof of (IdII) in a former note :

                D Ù Rab := d Ù Rab + ωac Ù Rcb − Rac Ù ωcb = 0 .                                 (A6)


[1] S.M. Carroll, Spacetime and Geometry, Addison Wesley 2004, ISBN 0-8053-8732-3