 ## On the Hodge dual of the first Bianchi identity

### Gerhard W. Bruhn, Darmstadt University of Technology

Dec 3, 2008

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

(1)         D Ù Ta = Rab Ù qb .

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

(1~)         D Ù Ta~ = Rab~ Ù qb

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)         Dμ Tκμν = Rκμμν

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 .

### 1. The unit-2-sphere S˛ in Rł

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

(1.1)         ds˛ = dθ˛ + sin˛θ dφ˛ = dx1 2 + sin˛ x1 dx2 2

with the metric tensors

(1.2)                 (gμν) = diag( 1, sin˛ x1) ,         (gμν) = diag( 1, 1/sin˛ x1) .

using the numbering of indices

(1.3)         1 ~ θ, 2 ~ φ

There are only a few non-vanishing Christoffel coefficients

(1.4)         Γ122 = − sin x1 cos x1 ,     Γ212 = Γ221 = cot x1 ,

while all other Christoffels Γκμν vanish.

The torsion Tκ is given by

(1.5)         Tκμν = Γκμν − Γκνμ = 0 ,

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

### 2. The Riemann tensor of S˛

The Riemann tensor is given by

(2.1)         Rκλμν = ∂μΓκλν − ∂νΓκλμ + ΓκμρΓρνλ − ΓκνρΓρμλ

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

(2.3)                 Rκλμν = 0     if     μ = ν.

### 3. The check

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

(3.1)         0 = Rκμμν = Rκμαβ gμα gνβ .

Therefore, due to the diagonal form (1.2) of (gμρ), we have to check:

(3.2)         (Rκ11β g11 + Rκ22β g22) gνβ = (Rκ111 g11 + Rκ221 g22) gν1 + (Rκ112 g11 + Rκ222 g22) gν2 .

This is:

for ν=1:     Rκμμ1 = (Rκ111 g11 + Rκ221 g22) g11 + 0 = (Rκ111 g11 + Rκ221 g22) g11 = Rκ221 g22 g11 ,

for ν=2:     Rκμμ2 = 0 + (Rκ112 g11 + Rκ222 g22) g22 = (Rκ112 g11 + Rκ222 g22) g22 = Rκ112 g11 g22 ,

i.e. the test reduces to:

(3.3)         Rκ221 = 0 ?         and         Rκ112 = 0 ?

We consider the special case κ = 1 to obtain:

(3.4)         R1221 = 0 ?         and         R1112 = 0 ?

The check 'R1221 = 0 ?' means in detail

(3.5)         R1221 = ∂2Γ121 − ∂1Γ122 + (Γ121Γ112 + Γ122Γ212) − (Γ111Γ122 + Γ112Γ222) = 0 ?
=o                             =o                               =o                 =o

thus

(3.6)         R1221 = − ∂1Γ122 + Γ122Γ212 = − sin˛ x1 ≠ 0 .

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

### Eq.(2) is invalid in general.

Remark: Another counter example to eq.(2) is given by the Schwarzschild metric in : Sect.1.1.4 gives symmetric Christoffel connection, hence the torsion is zero. However, due to Sect.1.1.12 we have Roμμo ≠ 0, again contradicting eq.(2).

### References

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

 M.W. Evans, H. Eckardt, Spherically symmetric metric with perturbation a/r
http://www.atomicprecision.com/blog/wp-filez/a-r.pdf

 S.M. Carroll, Lecture Notes on General Relativity, p.60 f.,