Dec 3, 2008

One of the basic equations of differential geometry is the 1st Bianchi identity

(1)
D Ù T^{a} =
R^{a}_{b}
Ù q^{b} .

In their webnote [1] the authors attempt to dualize that identity by claiming the Hodge dual of eq.(1) to be

(1^{~})
D Ù T^{a~} =
R^{a}_{b}^{~}
Ù q^{b}

as an up to now unknown further *basic equation of differential geometry.*

To prove that new equation (1^{~}) is transformed to the tensorial equation

(2)
D_{μ} T^{κμν} =
R^{κ}_{μ}^{μν}

which is considered to be *equivalent* to the dualized 1st
Bianchi identity (1^{~}).
Therefore the tensorial equation (2) should be valid for arbitrary sample
manifolds of Riemannian differential geometry.
We'll check this assertion (2) by an elementary example below.
The reader will find a much deeper treatment of the topic by W.A. Rodrigues Jr.
in [4].

Using some basic informations from S.M. Carroll [3] we have the metric

(1.1)
ds² = dθ² + sin²θ dφ² =
dx^{1 2} +
sin² x^{1} dx^{2 2}

with the metric tensors

(1.2)
(g_{μν}) = diag( 1, sin² x^{1}) ,
(g^{μν}) = diag( 1, ^{1}/_{sin² x1}) .

using the numbering of indices

(1.3) 1 ~ θ, 2 ~ φ

There are only a few non-vanishing Christoffel coefficients

(1.4)
Γ^{1}_{22} =
− sin x^{1} cos x^{1} ,
Γ^{2}_{12} =
Γ^{2}_{21} =
cot x^{1} ,

while all other Christoffels Γ^{κ}_{μν} vanish.

The torsion T^{κ} is given by

(1.5)
T^{κ}_{μν} =
Γ^{κ}_{μν} −
Γ^{κ}_{νμ} = 0 ,

i.e. vanishing due to the symmetry of the Christoffels in their lower indices μ,ν.

The Riemann tensor is given by

(2.1)
R^{κ}_{λμν} =
∂_{μ}Γ^{κ}_{λν} −
∂_{ν}Γ^{κ}_{λμ} +
Γ^{κ}_{μρ}Γ^{ρ}_{νλ}
−
Γ^{κ}_{νρ}Γ^{ρ}_{μλ}

being antisymmetric in μ,ν as is well-known. Therefore we have especially

(2.3)
R^{κ}_{λμν} = 0
if
μ = ν.

Due to the vanishing of torsion (1.5) the equation (2) to be checked reduces to

(3.1)
0 =
R^{κ}_{μ}^{μν} =
R^{κ}_{μα}_{β}
g^{μα}
g^{νβ} .

Therefore, due to the diagonal form (1.2) of (g^{μρ}),
we have to check:

(3.2)
(R^{κ}_{11}_{β} g^{11} +
R^{κ}_{22}_{β} g^{22})
g^{νβ}
=
(R^{κ}_{111} g^{11} +
R^{κ}_{221} g^{22})
g^{ν1} +
(R^{κ}_{112} g^{11} +
R^{κ}_{222} g^{22})
g^{ν2} .

This is:

for ν=1:
R^{κ}_{μ}^{μ1}
=
(R^{κ}_{11}_{1} g^{11} +
R^{κ}_{22}_{1} g^{22})
g^{11} + 0
=
(R^{κ}_{11}_{1} g^{11} +
R^{κ}_{22}_{1} g^{22})
g^{11}
=
R^{κ}_{22}_{1} g^{22}
g^{11} ,

for ν=2:
R^{κ}_{μ}^{μ2}
=
0 +
(R^{κ}_{112} g^{11} +
R^{κ}_{222} g^{22})
g^{22}
=
(R^{κ}_{112} g^{11} +
R^{κ}_{222} g^{22})
g^{22}
=
R^{κ}_{112} g^{11}
g^{22} ,

i.e. the test reduces to:

(3.3)
R^{κ}_{221} = 0 ?
and
R^{κ}_{112} = 0 ?

We consider the special case κ = 1 to obtain:

(3.4)
R^{1}_{221 } = 0 ?
and
R^{1}_{112} = 0 ?

The check 'R^{1}_{221} = 0 ?'
means in detail

(3.5)
R^{1}_{221} =
∂_{2}Γ^{1}_{21} −
∂_{1}Γ^{1}_{22} +
(Γ^{1}_{21}Γ^{1}_{12} +
Γ^{1}_{22}Γ^{2}_{12}) −
(Γ^{1}_{11}Γ^{1}_{22} +
Γ^{1}_{12}Γ^{2}_{22}) = 0 ?

=o
=o
=o
=o

thus

(3.6)
R^{1}_{221} = −
∂_{1}Γ^{1}_{22} +
Γ^{1}_{22}Γ^{2}_{12}
=
− sin² x^{1}
≠ 0 .

Therefore we have obtained a **negative check result**:
The test equation (2) is not fulfilled
for the unit-2-sphere *S*² which means:

**Remark**:
Another counter example to eq.(2) is given by the Schwarzschild metric in [2]:
Sect.1.1.4 gives symmetric Christoffel connection, hence the torsion is zero.
However, due to Sect.1.1.12 we have
R^{o}_{μ}^{μo} ≠ 0, again contradicting eq.(2).

[1] M.W. Evans, H. Eckardt, *Violation of the Dual Bianchi Identity
by Solutions of the Einstein Field Equation*

Violation of the Dual Bianchi Identity

[2] M.W. Evans, H. Eckardt, *Spherically symmetric metric with perturbation a/r*

http://www.atomicprecision.com/blog/wp-filez/a-r.pdf

[3] S.M. Carroll, *Lecture Notes on General Relativity*, p.60 f.,

http://www2.mathematik.tu-darmstadt.de/~bruhn/Carroll84-85.bmp

[4] W.A. Rodrigues Jr., *Differential Forms on Riemannian (Lorentzian) and
Riemann-Cartan Structures and Some Applications to Physics*

Ann. Fond. L. de Broglie

http://arxiv.org/pdf/0712.3067

[5] G.W. Bruhn, *Evans' Duality Experiments*

http://www2.mathematik.tu-darmstadt.de/~bruhn/Duality.html