16.03.07: Wrong factors 4 and ¼ removed from Eqs.(14.21') and (14.22') respectively

A quotation from Evans' blog site

*"Cartan geometry is available in a standard and well known computer package such as Maple,
a fact which shows that the attacks on Cartan geometry are pseudo-mathematics.
Cartan geometry is actually wired into the circuits, and this computer code can be used to
automate calculations. This process is self checking and self consistent. British Civil List Scientist."*

*"After all the major parts are entered, we can take all rebuttals and criticism from Bruhn, Lakhtakia etc.
and show where their faulty logic is. After this is compiled, we can send a letter to the main Wikipedia editors
along with Myron’s list of papers and books and his copies of Civil List appointment and medals etc. along
with the rebuttals with solutions and examples demonstrating their correctness.
*

*Best regards,
*

*Ross Pawley
"*

So here are some pseudo-mathematical attacks on "Cartan geometry" **by M.W. Evans himself**
that he should (let) check by Maple. To recognize these errors other people don't need Maple,
it belongs to their standard knowledge:

p. 262 of Evans' GCUFT book Vol. 1:

Using ... the Cartan convention [11]

q^{a}_{μ} q^{μ}_{a} = 1 ,
(14.20)

p.289 of Evans' GCUFT book Vol. 1:

Using the Cartan convention [13]

q^{a}_{μ} q^{μ}_{a} = 1 ,
(15.80)

Again p.261 f. of Evans' GCUFT book Vol. 1:

R_{μν} = R^{a}_{μ} q^{b}_{ν}
η_{ab},
(14.13)

. . .

together with the definitions (13) (= eq. (14.13)) . . . we obtain:

R = g^{μν}R_{μν}

= q_{μ}^{a} q^{b}_{ν}
η^{ab}R^{a}_{μ}
q^{b}_{ν}
η_{ab}
(14.21)

= (η^{ab}η_{ab})(q^{ν}_{b}
q^{b}_{ν})(q_{μ}^{a} R^{a}_{μ}) =
4 q_{μ}^{a} R^{a}_{μ} .

a) Show by Maple or anyhow that (14.21) is an invalid equation.

b) The correct equation as a consequence of eq. (14.13) is

R = q_{a}^{μ} R^{a}_{μ} .
(14.21')

Confirm that by Maple.

Multiply either side of Eq. (21) (= eq. (14.21)) by
q^{a}_{μ} to obtain

R^{a}_{μ} =
¼ Rq^{a}_{μ} ,
(14.22)

This should according to the correct Eq.(14.21') read

R^{a}_{μ} =
R q^{a}_{μ} ,
(14.22')

c) Show by Maple that eq. (14.21') (the correction of eq. (14.21)) cannot be resolved for
the 16 quantities R^{a}_{μ}. Thus, eq. (14.22) is invalid.

**Hint** Confirm by maple that
eq. (14.21') is fulfilled by

R^{a}_{μ} = R q^{a}_{μ} + q^{b}_{μ}
where b=b(a):=(a+1) mod 4.

p.317 of Evans' GCUFT book Vol. 1:

Multiplying both sides of Eq (B.3) by q^{b}_{λ}
and using:

q^{b}_{λ}q^{λ}_{b} = **1**
(B.4)

p.319 of Evans' GCUFT book Vol. 1:

In order to show that this is zero use:

q^{λ}_{c}q^{c}_{λ} = **1**
(B.25)

quote from p.55:

The symmetric metric tensor is then defined through the line element, a one form of differential geometry:

ω_{1} = ds² = q^{ij(S)}du_{i}du_{j} ,
(3.12)

and the anti-symmetric metric tensor through the area element, a two form of differential geometry:

ω_{2} = dA =
− ½ q^{ij(A)}du_{i}Ùdu_{j} .
(3.13)

These results generalize as follows to the four dimensions of any non-Euclidean space-time:

ω_{1} = ds² = q^{μν(S)}du_{μ}du_{ν},
(3.14)

ω_{2} = *ω_{1} = dA =
− ½ q^{μν(A)}du_{μ}Ùdu_{ν}.
(3.15)

Try to compute *ω_{1} with Maple. The Hodge dual operator * is only defined for
antisymmetric forms. Is the form ω_{1} antisymmetric?

In differential geometry the element du_{σ} is dual to the wedge product
du_{μ}Ùdu_{ν}.

So try to prove the claim

Again from the same page:

The symmetric metric tensor is:

[ h_{0}² h_{0}h_{1} h_{0}h_{2} h_{0}h_{3} ]

[ h_{1}h_{0} h_{1}² h_{1}h_{2} h_{1}h_{3} ]

q^{μν(S)} = [
]
(3.16)

[ h_{2}h_{0} h_{2}h_{1} h_{2}² h_{2}h_{3} ]

[ h_{3}h_{0} h_{3}h_{1} h_{3}h_{2} h_{3}² ]

Compute by Maple the determinant of the matrix [q^{μν(S)}] and - if possible -
its inverse [q^{μν(S)}]^{−1}.
The result will be a vanishing determinant as we have all line vectors being parallel: Each line is parallel to the vector
[ h_{0} h_{1} h_{2} h_{3} ]. Thus, the matrix is singular having no inverse.
However, the metric must have a nonsingular matrix, since
its inverse is needed in tensor calculus too.

Let ds² be defined by eq. (3.14) and the coefficients q^{μν(S)} be given by eq. (3.16).
All quantities are assumed reals. Prove with Maple the identiy

ds² = h_{μ}h_{ν}du^{μ}du^{ν} =
(h_{μ}du^{μ})^{2} ≥ 0 .

(We have written the indices of the differentials du in the upper position as is use in tensor calculus.)

Compare with the Minkowski-metric (q_{μν}) := diag(−1, +1,+1,+1) which
yields *timelike* directions if
du^{0}^{2} >
du^{1}^{2}+du^{2}^{2}+du^{3}^{2}
and *spacelike* directions if
du^{0}^{2} <
du^{1}^{2}+du^{2}^{2}+du^{3}^{2} .
Therefore equ. (3.16) cannot be used to define the (flat) Minkowski metric.