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)

Not only that Evans attributes his wrong equation to Cartan he even falsely refers to S.M. Carroll's Lecture Notes, where - of course - nothing of that nonsense can be found.

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^μνR_μν                                                                                                      
= q_μ^a q^b_ν η^abR^a_μ q^b_ν η_ab                                                                 (14.21)
= (η^abη_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.

Thus, eq. (14.22) is no implication of eq. (14.21').

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

Multiplying both sides of Eq (B.3) by q^b_λ and using:

q^b_λq^λ_b = 1                                                                 (B.4)

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

In order to show that this is zero use:

q^λ_cq^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:

ω₁ = ds² = q^ij(S)du_idu_j ,                                                                 (3.12)

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

ω₂ = dA = − ½ q^ij(A)du_iÙdu_j .                                                                 (3.13)

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

ω₁ = ds² = q^μν(S)du_μdu_ν,                                                                 (3.14)

ω₂ = *ω₁ = dA = − ½ q^μν(A)du_μÙdu_ν.                                                                 (3.15)

Try to compute *ω₁ with Maple. The Hodge dual operator * is only defined for antisymmetric forms. Is the form ω₁ antisymmetric?

In differential geometry the element du_σ is dual to the wedge product du_μÙdu_ν.

So try to prove the claim

*du_σ = du_μÙdu_ν

Again from the same page:

The symmetric metric tensor is:

[ h₀² h₀h₁ h₀h₂ h₀h₃ ]                                                                         
[ h₁h₀ h₁² h₁h₂ h₁h₃ ]                                                                         
q^μν(S)   =   [                              ]                                                                (3.16)
[ h₂h₀ h₂h₁ h₂² h₂h₃ ]                                                                         
[ h₃h₀ h₃h₁ h₃h₂ h₃² ]                                                                         

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 [ h₀ h₁ h₂ h₃ ]. 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.

Thus, eq. (3.16) cannot provide a metric.