 Last Update: Sept 18, 2005, 08:00 pm

## Evans Proving the Metric Compatibility?

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

The definition of covariant derivatives is not uniquely determined. S.M. Carroll  used the well known tetrad identity

(1)                                                 ∂ρ qaμ + ωaρs qsμ − Γσρμ qaσ = 0

to define a covariant derivative Ñρ that admits interpreting of (1) as Ñρ qaμ = 0 : Carroll defines

(2)                                                 Ñρ qaμ := ∂ρ qaμ + ωaμs qsρ − Γσρμ qaσ .

In the following we'll speak here of Ñρ as of the "Carroll derivative" for matter of distinction.

Using the Carroll derivative we have the "tetrad postulate"

(3)                                                                 Ñρ qaμ = 0

The rule: Each upper Latin index (e.g. at qa· ) causes an additive term + ωaμb qb· , while a lower Latin index (e.g. at q·a) gives rise to the additive term − ωaμb q·a. Greek indices have to be treated as usual [2; (3.1),(3.12)]

Example:

(4)                                                         Ñρ qμa := ∂ρ qμa − ωsρa qμs + Γμρσ qσa .

Let η = (ηab) := diag(−1, 1, 1, 1) be the Minkowski matrix. Then due to the constancy of each ηab we obtain

(5)                 Ñρ ηab = ∂ρηab − ωsρa ηsb − ωsρb ηas = − ωsρa ηsb − ωsρb ηas = − ωb,ρa − ωa,ρb .

### Note that the Carroll derivative of the constant ηab does NOT vanish!

M.W. Evans applies the Carroll derivative in the following way: He wants to calculate Ñρ gμν by referring to the equation gμν = ηab qaμ qbν. Using the Leibniz rule Evans concludes due to (3)

(6)                                 Ñρgμν = Ñρab qaμ qbν) = (Ñρηab) qaμ qbν + 0 + 0 .

Here he erroneously assumes the derivative Ñρ of a constant term ηab to be ZERO (also - among other errors - e.g. on [1; p.46 and p.70] where the Minkowski matrix is unwritten to have an elegant notation) Ñρηab = 0, which would prove the metric compatibility from the tetrad identity.

However, the correct result is due to (4)

(7)                           Ñρgμν = Ñρab qaμ qbν) = − (ωb,ρa + ωa,ρb) qaμ qbν = − (ων,ρμ + ωμ,ρν)

which differs from zero in general.

Evans should have recognized his conclusion to be wrong, since one can get the desired result directly also by evaluating :

(8)                                                 Ñρgμν = ∂ρgμν − Γλρμgλν − Γλρνgμλ

where the term ∂ρgμν = ∂ρab qaμ qbν) is to be calculated with the Leibniz rule of partials:

Ñρ gμν = ∂ρ gμν − Γλρμ gλν − Γλρν gμλ
= ηab (∂ρqaμ) qbν + ηab qaμ (∂ρqbν) − Γλρμ gλν − Γλρν gμλ
= [ηab (∂ρqaμ) qbν − Γλρμ gλν] + [ηab qaμ (∂ρqbν) − Γλρν gμλ]
= ηab (∂ρqaμ − Γλρμqaλ) qbν + ηab qaμ (∂ρqbν − Γλρν qbλ)

By using the tetrad identity (1) we obtain finally

Ñρ gμν = ηab [(∂ρqaμ − Γλρμqaλ) qbν + qaμ (∂ρqbν − Γλρν qbλ)]
= − ηabaρc qcμ qbν + ωbρc qaμ qcν)
= − ηab ωaρc (qcμ qbν + qbμ qcν)
= − ωaρc (qcμ ηab qbν + ηabqbμ qcν)
= − ωaρc (qcμ qσa gσν + qcν qσa gσμ)
= − ωaρc qσa (qcμ gσν + qcν gσμ)
= − ωσρc (qcμ gσν + qcν gσμ)
= − (ωσρμ gσν + ωσρν gσμ)
= − (ων,ρμ + ωμ,ρν).

in accordance with Equ.(7). Therefore we have

### The tetrad postulate (3) does not imply the metric compatibility of the connection.

Remark Evans reference  to the covariant derivative to be used is ambiguous. S.M. Carroll gives two different definitions: The "Carroll definition" at [2; p.91] and the "usual" one for D at [2; p.56 (3.1) and p.58 (3.12)]. At [1; p.46 (2.182) we find the application of the usual definition [2; p.56 (3.1)]

(4')                                                         Dν qμa := ∂ν qμa + Γμνλ qλa

in Evans' notation with suppressed Latin indices

Dν qμ := ∂ν qμ + Γμνλ qλ                                                         (2.182)

With the extended equation (4') instead of (2.182) we obtain

Dρ gμν = Dρab qμaqνb) = ηab [(Dρqaμ) qbν + qaμ (Dρqbν)]
= ηab [(∂ρ qμa + Γμρλ qλa) qνb + qμa (∂ρ qνb + Γνρλ qλb)].

Here the analoguous equation to the tetrade identity (1), the equation

(1')                                                 ∂ρ qμa − ωsρa qμs + Γμρσ qσa = 0

can be used to obtain

= − ηabsρa qμs qνb + ωsρb qνs qμa) = − ηabμρa qνb + ωνρb qμa)
= − (ωμρν + ωνρμ) .

The reader is kindly asked to prove the analoguous equation

(7')                                                           Dρgμν = − (ων,ρμ + ωμ,ρν),

which agrees with (7) though different covariant differential operators Ñρ and Dρ were used.

### References

    M.W. Evans, Generally Covariant Unified Field Theory, the geometrization of physics,