Open System for Geniuses - by Chronostalker

1 July, 2005

Generally Covariant Unified Field Theory - Myron W. Evans

Filed under: General

Browsing through different websites lately, I noticed a new web page on quantumfuture.net - the page is intended to deal with a new book by Myron W. Evans entitled: "Generally Covariant Unified Field Theory".

Looking at the book, reading it here or there, I realized that Myron W. Evans deserves as much attention, if not more, as William Tiller or Jack Sarfatti. Therefore I have undertaken the pain of reviewing in detail some first few pages of the book - namely the pages dealing with what Evans calls "Mathematical Prerequisite".

It is clear that Evans is yet another "impressionist" among physicists and chemists. From an impressionist you would not expect knowledge or understanding of the mathematical concepts he or she may be using - that would be too much. It is up to you to make sense of what is written - it may inspire you to produce something of your own. And there is nothing wrong with this category of writers. What is important is that the reader knows in advance that the masterpiece does not belong to the "realistic" category. That it is much like a game of free association and of science-fiction.

It is interesting that Evans apparently submitted his masterpiece to Springer, and that it was rejected by Springer - after some interventions from "realists". It is also interesting that the Chief Editor of Foundations of Physics, Alwyn van der Merwe, has problems with distinguishing between impressionism and realism as I will soon show.

I have to add that even if the part of the book reviewed below is plagued with errors and provides ample evidence that the author has rather poor understanding of elementary mathematical concepts such as “vector, straight line, metric, differential form, tensor, matrix”, and that no one has read the text if only to correct evident and elementary calculational mistakes and contradicting formulas - nevertheless that does not mean that all the ideas in the book are wrong. Some may be even right - but it would be a pity if this particular book or this particular author were permitted to discredit the idea of "non-symmetric metric" or "unified field theory". There are, for instance, "idiots savant" - where an "idiot savant" is defined as " a person with autism that has extraordinary skills in certain domains in spite of cognitive deficiencies in most others."

Anyway, here is the first part of the "Referee Report" - in case someone looks for such. (Note: Valuable comments on in the same spirit, dealing with errors in the papers of Myron W. Evans, can be also found on the web site of Gerhard W. Bruhn at the University of Darmstadt)


Referee’s Report on "Generally Covariant Unified Field Theory by Myron W. Evans"

1 Basic Geometrical Concepts

Part I of the book is entitled “Mathematical Prerequisite”. The we have:

1.1 Basic Definitions

We expect therefore that the content will be mathematically precise, and that is what I will be looking at while commenting upon the text. Evans writes:
In this first section we define the basic concepts needed for this task [1-5]. The metric is developed from the first principles of curvilinear coordinate analysis [6]. We start with basic definitions. First consider the straight line in three-dimensional space,


$\displaystyle {\bf r} = x{\bf i} + y{\bf j} + z{\bf k}.$ (1.1)

Comment: There is no explanation what are $ {\bf i},{\bf j},{\bf k}.$ There is no explanation of what kind of `three-dimensional space’ we are supposed to be in. Therefore, when things are not defined, the reader has to guess. Here our guess is that we are in $ \mathbb{{\bf R}}^3$ endowed with the standard Euclidean metric, and that $ {\bf i},{\bf j},{\bf k}$ are the standard unit vectors. But even so $ {\bf r} = x{\bf i} + y{\bf j} + z{\bf k}$ is not a straight line. If $ x,y,z$ are constant real numbers, then $ {\bf r}$ is a vector, not a `straight line’. On the other hand, if $ x,y,z$ are variable, then $ {\bf r}$ goes over all the space - not over a straight line.

Then we read:

The unit vectors are defined as:


$\displaystyle {\bf i}= \frac{\partial {\bf r}}{\partial x}/\left\vert\frac{\partial {\bf r}}{\partial x}\right\vert \dots$ (1.2)

and the metric vectors as


$\displaystyle {\bf q}_x=\left\vert\partial{\bf r}/\partial x\right\vert{\bf i}, \ldots$ (1.3)

Comment: That does not make any sense. First of all, if $ {\bf r}$ is defined through $ {\bf i},{\bf j},{\bf k},$ as in Eq. (1.1) then $ {\bf i},{\bf j},{\bf k}$ must not be defined in terms of $ {\bf r}$ as in Eq. (1.2). Eq. (1.2) is then a useless tautology. It is clear now that the author can not distinguish between a definition and a tautology . Another unnecessary definition is to define $ {\bf q}_x$ by $ {\bf i}.$ What’s the point to introduce a new notation for the same object. Perhaps Evans wants to introduce a vector-valued one-form, or an “orthonormal frame”, but then he should say so. Then Evans writes:
The metric element is


$\displaystyle q_x=\vert\partial {\bf r}/\partial x \vert=1=q_y=q_z,$ (1.4)

and the line element is defined as


$\displaystyle d{\bf r}=\frac{\partial {\bf r}}{\partial x}dx+\frac{\partial {\bf r}}{\partial y}dy+\frac{\partial {\bf r}}{\partial z}dz.$ (1.5)

Comment: Now, in mathematics, when we speak about the line element , we usually mean $ ds^2$ of the Riemannian geometry. So, for instance, Wikipedia defines:
“The line element in mathematics can most generally be thought of as the square of the change in a position vector in an affine space equated to the square of the change of the arc length.”
It is evident that Evans has something else in mind. But he does not tell us what. If he means simply $ d{\bf r}$ - then it should be called $ d{\bf r}$ - not a “line element”.
The metric tensor is


$\displaystyle q_{ij} = g_{ji} = \frac{\partial{\bf r}}{\partial x}\cdot\frac{\partial{\bf r}}{\partial y}, $   etc. (1.6)

If $ g_{ij} = 0$ for $ i\neq j$ then the coordinate system is orthogonal.

Comment: Here we see that someone evidently was typesetting Evan’s manuscript, and that the result was not proofread. It should read $ q_{ij} = 0$ for $ i\neq j$ rather than $ g_{ij} = 0$ for $ i\neq j$ . Indeed, the letter $ q$ is very similar to the letter $ g$ and most probably there should be $ g_{ij} = g_{ji}$ or $ q_{ij} = q_{ji}$ rather than $ q_{ij} = g_{ji}.$ But if he decides to use $ q_{ij}$ for the metric tensor, then the next line should read $ q_{ij} = 0$ for $ i\neq j$ rather than $ g_{ij} = 0$ for $ i\neq j.$ This is a minor issue though. We see from the above that our initial guess that $ {\bf i},{\bf j},{\bf k}$ denote the standard basis of the Euclidean space $ {\bf R}^3$ was not correct because now $ {\bf i}$ and $ {\bf j}$ need not be mutually orthogonal. Yet, for some unspecified reason, they happen to be of unit length. This is not a good start for a chapter entitled “Mathematical Prerequisite”, but let’s continue. Perhaps it will get better. Evans writes:
If we consider the functional relations that define the complex circular basis [1-5] of three-dimensional space,


$\displaystyle e^{(1)} = (1/ \sqrt{2})(x - iy),\quad e^{(2)} = (1/\sqrt{2})(x + iy),\quad e^{(3)} = z$ (1.7)

then


$\displaystyle x = (1/ \sqrt{2})(e^{(1)} + e^{(2)}),\quad y = (i/ \sqrt{2}) (e^{(1)} - e^{(2)}),\quad z = e^{(3)}$ (1.8)

are curvilinear coordinate relations in this three-dimensional space. The curve (1.1) can therefore be written as


$\displaystyle {\bf r}= (1/ \sqrt{2}) (e^{(1)} + e^{(2)}) {\bf i}+ (i/ \sqrt{2}) (e^{(1)} - e^{(2)}){\bf j}+ e^{(3)}{\bf k},$ (1.9)

giving the three unit vectors in the complex circular basis as


$\displaystyle {\bf e}^{(1)}=\frac{\partial {\bf r}}{\partial e^{(1)}}/\left\... ...ac{\partial {\bf r}}{\partial e^{(1)}}\right\vert=1/\sqrt{2}({\bf i}-i{\bf j})$    


$\displaystyle {\bf e}^{(2)}=\frac{\partial {\bf r}}{\partial e^{(2)}}/\left\... ...ac{\partial {\bf r}}{\partial e^{(2)}}\right\vert=1/\sqrt{2}({\bf i}+i{\bf j})$ (1.10)


$\displaystyle {\bf e}^{(3)}=\frac{\partial {\bf r}}{\partial e^{(3)}}/\left\vert\frac{\partial {\bf r}}{\partial e^{(3)}}\right\vert={\bf k}$    

Comment: First of all notice that Eq (1.7) does not define a basis. It defines a complex-valued coordinate $ e^{(1)}$ instead of two real coordinates $ x,y.$ Then $ e^{(2)}$ is nothing but the complex conjugate of $ e^{(1)}.$ Not only we do not have a basis here, we do not have any circularity in Eq (1.7) - the relations there are linear. We are not being told the point of introducing a complex coordinate at this point. We are not being told the point of introducing a complex coordinate at this point.. Moreover, contrary to what is written right after Eq (1.8), $ x,y,z$ are not “curvilinear coordinate relations” as there is nothing curvilinear there. These are all linear relations. Moreover Eq (1.9) is not a curve, similar as Eq. (1.1) is not a straight line. It is evident that neither a straight line, a curve, nor a basis is understood by the author. If he does understand them, then he is not being careful in his writing about the proper use of these mathematical concepts. Furhermore, the (complex) vectors $ {\bf e}^{(1)},{\bf e}^{(2)}$ are not unit vectors, unless $ {\bf i},{\bf j}$ are mutually orthogonal, but just a while ago they were not assumed to be necessarily orthogonal. Finally, it becomes evident that no one ever bothered to read the text with attention. The formula above is trivially wrong. Indeed, computing $ \partial{\bf r}/\partial e^{(1)}$ from the formula (1.9) we find

$\displaystyle \partial{\bf r}/\partial e^{(1)}=1/\sqrt{2}({\bf i}+i{\bf j})$

and

$\displaystyle \partial{\bf r}/\partial e^{(2)}=1/\sqrt{2}({\bf i}-i{\bf j}).$

Therefore Eq (1.10) should read:


$\displaystyle {\bf e}^{(1)}=\frac{\partial {\bf r}}{\partial e^{(1)}}=1/\sqrt{2}({\bf i}+i{\bf j})$    


$\displaystyle {\bf e}^{(2)}=\frac{\partial {\bf r}}{\partial e^{(2)}}=1/\sqrt{2}({\bf i} -i{\bf j}) \qquad (1.10a)$    


$\displaystyle {\bf e}^{(3)}=\frac{\partial {\bf r}}{\partial e^{(3)}}={\bf k}$    


where we skip the meaningless divisions by $ 1.$
Note We are differentiating a function, in our case $ {\bf r},$ of a complex variable. In complex analysis a complex variable, such as $ {\bf e}^{(1)},$ and its complex conjugate, such as $ {\bf e}^{(2)}$ are considered as independent, and differentiating is usually understood in terms of local power series decomposition.
In this basis the line element is


$\displaystyle {\bf r}=\frac{\partial{\bf r}}{\partial e^{(1)}} /\left\vert\fra... ...artial e^{(3)}}/\left\vert\frac{\partial{\bf r}}{\partial e^{(3)}}\right\vert$ (1.11)

and the metric vectors are


$\displaystyle {\bf q}^{(1)}=\partial{\bf r}/\partial e^{(1)}={\bf e}^{(1)},\qua... ...f e}^{(2)},\quad {\bf q}^{(3)}=\partial{\bf r}/\partial e^{(3)}={\bf e}^{(3)}.$ (1.12)

These vectors form the $ O(3)$ symmetry cyclic relations


$\displaystyle {\bf q}^{(1)}\times {\bf q}^{(2)}=i{{\bf q}^{(3)*}}, {\bf q}^{... ...}^{(3)}=i{\bf q}^{(1)*}, {\bf q}^{(3)}\times {\bf q}^{(1)}=i{{\bf q}^{(2)*}}$ (1.13)

where $ *$ denotes complex conjugation.

Comment: The last line has wrong signs. Taking into account the previous errors in calculating $ \partial{\bf r}/\partial e^{(1)}$ $ \partial {\bf r}/\partial e^{(2)}$ from Eq. (1.9) we should have minus signs there, i.e.,


$\displaystyle {\bf q}^{(1)}\times {\bf q}^{(2)}=-i{{\bf q}^{(3)*}}, {\bf q}^{... ...^{(1)*}, {\bf q}^{(3)}\times {\bf q}^{(1)}=-i{{\bf q}^{(2)*}},\qquad (1.13a)$    

- unless it was the equations (1.7) and (1.8) that were wrong! It is impossible to say, when the author writes two consecutive formulas, one contradicting the other, which one is an error and which one is a typo. It is possible, however, to deduce that no one read the text with care.

We continue with GCUFT:

Consider the Cartesian unit vector system $ {\bf i},{\bf j},{\bf k}.$ The metric tensor is formed from


$\displaystyle {\bf q}_x={\bf i},\quad {\bf q}_y={\bf j}, {\bf q}_z={\bf k}$ (1.14)

and is given by


$\displaystyle g_{11}={\bf q}_x\cdot{\bf q}_x={\bf i}\cdot{\bf i}=1,$    


$\displaystyle g_{22}={\bf q}_y\cdot{\bf q}_y={\bf j}\cdot{\bf j}=1,$    


$\displaystyle g_{33}={\bf q}_z\cdot{\bf q}_z={\bf k}\cdot{\bf k}=1,$ (1.15)


$\displaystyle g_{ij}=\delta_{ij}.$    

This symmetric metric tensor with unit diagonal coefficients represents an orthogonal coordinate system in flat, Euclidean space. It is formed from the dot products of the unit vectors $ {\bf i},{\bf j},{\bf k},$ because these unit vectors are the same as the metric vectors (Eq. (1.13)).

Comment: In differential geometry there is no such thing as “metric vectors”. Given a coordinate system $ x^i$ , the metric tensor is always built, by the very definition, out of scalar products of vectors $ \partial_i=\partial/\partial x^i$ tangent to the coordinates lines:

$\displaystyle g_{\mu\nu}\doteq (\partial_\mu,\partial_\nu)=\partial_\mu\cdot\partial_\nu.$

More generally, given any basis (or “non-holonomic frame”) $ e_i,$ the metric in this frame is given by

$\displaystyle g_{ij}\doteq (e_i,e_j)=e_i\cdot e_j.$

It is possible to form an anti-symmetric metric tensor by considering the well-known O(3) symmetry of the cross product of unit vectors


$\displaystyle {\bf i}\times{\bf j}={\bf k}, {\bf j}\times{\bf k}={\bf i}, {\bf k}\times{\bf i}={\bf j}.$ (1.16)

Comment: Normally the metric tensor is symmetric. Albert Einstein proposed a unified field theory based on a generalization of the metric, so that it could have an antisymmetric part. Einstein’s theory is described in details in “Einstein’s unified field theory” by M.A. Tonnelat, Gordon and Breach, NY, 1966. There, on p. 15 we can find “Relations between the Symmetrical and Antisymmetrical Parts of the tensors $ g_{\mu\nu}$ and $ g^{\mu\nu}.$ “ For some reason this classical reference is not quoted in GCUFT. What a possible antisymmetric part of $ g_{ij}$ has to do with cross products of the basic orthonormal vectors $ {\bf i},{\bf j},{\bf k}$ - is not clear. Therefore, let’s continue.
This cyclic $ O(3)$ symmetry is also given by the unit Cartesian metric vectors defined in Eq. (1.3):


$\displaystyle {\bf q}_x\times{\bf q}_y={\bf q}_z, {\bf q}_y\times{\bf q}_z={\bf q}_x, {\bf q}_z\times{\bf q}_x={\bf q}_y .$ (1.17)

Comment: One wonders what is the point of repeating the same formula twice? According to the definition (1.3), $ {\bf q}_x={\bf i},{\bf q}_y={\bf j},{\bf q}_z={\bf k},$ so what is the point in writing the same formula twice but just using different letters for denoting the same objects?
and by the well-known $ O(3)$ rotation generator matrices [7]


$\displaystyle J_x:=\begin{pmatrix}0&0&0\cr0&0&-i\cr0&i&0\cr\end{pmatrix}, J_... ...r\end{pmatrix}, J_z:=\begin{pmatrix}0&-i&0\cr i&0&0\cr0&0&0\cr\end{pmatrix},$ (1.18)

where


$\displaystyle [J_i,J_j]=i\epsilon_{ijk},\quad C_{ijk}=i\epsilon_{ijk},$ (1.19)

and where $ \epsilon_{ijk}$ is the Levi-Civita symbol, or fully antisymmetric rank-three unit tensor. For the $ O(3)$ group the Levi-Civita symbol also gives the three $ O(3)$ group structure constants.

Comment: Strictly speaking $ \epsilon_{ijk}$ is not a tensor according to the mathematical definition of a tensor, where by a tensor we understand a geometrical object associated to the bundle of all linear frames. It is a tensor, however, for the $ SO(3)$ group, while for $ O(3)$ it is a “pseudo-tensor” - as it is not invariant under inversions (orthogonal transformations of determinant $ -1$ ). It is not clear what $ J_x,J_y,J_z$ are doing here, as these are Hermitian matrices, and as such they are not generators. To get generators, one needs to multiply them by $ i$ (or by $ -i$ , depending on the convention). Saying that the group $ O(3)$ has “three structure constants” is misleading. What are these three constants? The word “three” should be removed.
The three antisymmetric metric tensors


$\displaystyle {q^{ij}}_x:=iJ_x=\begin{pmatrix} 0&0&0\cr0&0&1\cr0&-1&0\cr\end{p... ...ix}, {q^{ij}}_y:=iJ_y=\begin{pmatrix} 0&0&-1\cr0&0&0\cr1&0&0\cr\end{pmatrix},$    


$\displaystyle {q^{ij}}_z:=iJ_z=\begin{pmatrix}0&1&0\cr -1&0&0\cr0&0&0\cr\end{pmatrix}$ (1.20)

are equivalent to the three metric vectors $ {\bf q}_x,{\bf q}_y,{\bf q}_z$ and therefore to the three unit vectors $ i,j,k$ and form the $ O(3)$ symmetry cyclic relations


$\displaystyle [{q^{ij}}_x,{q^{ij}}_y]=-{q^{ij}}_z,\quad [{q^{ij}}_y,{q^{ij}}_z]=-{q^{ij}}_x,\quad [{q^{ij}}_z,{q^{ij}}_x]=-{q^{ij}}_y.$ (1.21)

giving an $ O(3)$ symmetry basis set for the representation of three dimensional space.

Comment: This paragraph is sloppier than what we encountered already. First of all, we notice that in “vectors $ i,j,k$ ” whoever was typesetting the text has forgotten to use bold letters. It should read “vectors $ {\bf i},{\bf j},{\bf k}$ , as $ i,j,k$ stand for “indices”. Second, we do not have “three antisymmetric metric tensors” here. We do have three antisymmetric matrices though, and we have one tensor (or rather pseudo-tensor), namely $ \epsilon$ . It is necessary to distinguish between a “tensor” and a “matrix”. Tensor can be represented by a matrix, but this matrix representation changes with the change of basis vectors (in our case, it changes when we rotate basis vectors). There are special tensors whose matrix representation is constant. They are sometimes called “constant tensors”. For the proper rotation group $ SO(3)$ there are two tensors: Kronecker’s $ \delta_{ij}$ and Levi-Civita $ \epsilon_{ijk}.$ The second one is, in fact, constant for the whole special linear group . (The identity tensor $ \delta^i_j$ is constant for all general linear, not necessarily orthogonal or unimodular transformations). Hence, the term “The three antisymmetric metric tensors” is simply wrong and tells us that the author can not distinguish between a tensor and a matrix. Additionally, the word “metric” in the above is meaningless. Finally, the term “representation of a three dimensional space” is is also meaningless, as what we have above is a particular basis of the Lie algebra of the Lie group $ SO(3).$
The elements of the antisymmetric metric tensor and its equivalent metric vector are related in contravariant-covariant tensor notation [7] by


$\displaystyle q_k:=-\frac{1}{2}\epsilon_{ijk}q^{ij}$ (1.22)

thus identifying the metric vector as an axial vector.

Comment: The relation above has nothing to do whatsoever with “metric” vectors or “metric tensors”. $ \epsilon_{ijk}$ represents here the volume three-form, and in any number $ n$ of dimensions a volume form can be used to convert $ p$ -vectors into $ n-p$ -forms. In our case, in an oriented orthonormal basis, the volume form is represented by $ \epsilon_{i_1\ldots i_n}$. We can transform any $ p$ -vector $ v^{i_1\ldots i_p}$ into its dual $ (n-p)$ -form $ *v_{j_1\ldots j_{n-p}}$ using the standard formula


$\displaystyle (*v)_{j_1\ldots j_{n-p}}=\frac{1}{p!} v^{i_1\ldots i_p} \epsilon_{i_1\ldots i_pj_1\ldots j_{n-p}}.$    


The inverse transform is


$\displaystyle v^{i_1\ldots i_p}=\frac{1}{(n-p)!} \epsilon^{i_1\ldots i_p j_1\ldots j_{n-p}}(*v)_{j_1\ldots j_{(n-p)}}$    


When specialized to our case of $ n=3, p=2$ , we get


$\displaystyle q_k=\frac{1}{2}\epsilon_{ijk}q^{ij},\qquad q^k=\frac{1}{2}\epsilon^{ijk}q_{ij},\quad (1.22a)$    


$\displaystyle q^{ij}=\epsilon^{ijk}q_k,\quad q_{ij}=\epsilon_{ijk}q^k. \qquad (1.22b)$    


Evans’ formula (1.22) has a wrong sign. In fact, in the examples below, Evans evidently uses the standard textbook formulas (1.22a,1.22b) not his formula (1.22)!
In order to build up the antisymmetric tensor corresponding to $ {\bf k}= {\bf q}_z,$ we have


$\displaystyle \vert{\bf q}_z\vert=-q^3=\frac{1}{2}(\epsilon_{123}q^{12}+\epsilon_{213}q^{21}).$ (1.23)

The Levi-Civita elements are


$\displaystyle \epsilon_{123}=-\epsilon_{213}=1,$ (1.24)

so we obtain


$\displaystyle {q^{ij}}_z=-\begin{pmatrix}0&-q^{12}&0\cr q^{12}&0&0\cr 0&0&0\end{pmatrix}=-\begin{pmatrix}0&-1&0\cr 1&0&0\cr 0&0&0\end{pmatrix}.$ (1.25)

Comment In the paragraph above all is wrong except for the final results! First of al,l notice that Eq.(1.23) starts with $ \vert{\bf q}_z\vert=-q^3,$ which is rather silly, because this symbol denotes the length of the vector. While the author evidently wants to use the components here, the length has nothing to do with the standard relation between bi-vectors and one-forms (1.22a,1.22b) that is being discussed. Second, we see a mysterious object $ q^3$ which never appeared before, with a minus sign in front of it. What it is doing here and what does it mean - we can only guess. Probably Evans is fighting with himself trying to reconcile his wrong sign in (1.22) with the final answer (1.25) which is correct, as can be immediately seen from (1.22b).
The complete antisymmetric metric tensor corresponding to the sum $ {\bf i}+{\bf j}+{\bf k}$ is obtained from


$\displaystyle q_1=-\frac{1}{2}(\epsilon_{231}q^{23}+\epsilon_{321}q^{32}),\quad q_2=-\frac{1}{2}(\epsilon_{312}q^{31}+\epsilon_{132}q^{13}),$ (1.26)

and is given by


$\displaystyle q^{ij}=-\begin{pmatrix}0&-1&1\cr 1&0&-1\cr -1&1&0\end{pmatrix}.$ (1.27)

Comment: Again, what is above does not make sense except that the final result is right, but it is right when using our (1.22ab) not when using (1.22). Indeed, the vector $ {\bf i}+{\bf j}+{\bf k}$ has components $ (1,1,1).$ Next, because we are working in an orthonormal basis, there is no difference between covariant and contravariant components - they are the same. Therefore, according to (1.22b), the associated dual $ 2$ -form $ q_{ij}$ has components $ q_{ij}=\epsilon_{ij1}+\epsilon_{ij2}+\epsilon_{ij3}.$ That is,


$\displaystyle \left[q_{ij}\right]=\begin{pmatrix}0&1&-1\cr -1&0&1\cr 1&-1&0\end{pmatrix}.$ (1.28)


Let us stress that to call the two-form $ q_{ij}$ the complete antisymmetric metric tensor is misleading, and it is hard to imagine that this misleading is a result of a “chance”. Why would anybody call a two-form dual to some given vector a “totally antisymmetric metric tensor”? It is neither “totally” nor “metric”. It is simply “anti-symmetric”.
Similarly the anti-symmetric metric tensors corresponding to the metric vectors in the complex circular basis are given by


$\displaystyle q^{ij(1)}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&i\cr 0&0&1\cr -i&... ...j(2)}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&-i\cr 0&0&1\cr i&-1&0\end{pmatrix}.$ (1.29)

Comment: Again we have several options there as it is next to impossible for the reader to decide which of the contradictory statements of the author are to be taken seriously. As I already noticed above, the statements (1.9) and (1.10) contradict each other. They can not be true at the same time. One must be wrong. If we decide that (1.9) is wrong, so that (1.10) is correct, then (1.28) is correct provided however we calculate $ q^{ij(1)}$ and $ q^{ij(2)}$ using our formula (1.22b) rather than the wrong formula (1.22) of Evans. But again, using the term “anti-symmetric metric tensors corresponding to the metric vectors” instead of “dual two-forms” corresponding to vectors”, is inappropriate and misleading.
Therefore starting from the displacement vector (1.1) in Euclidean space-we have shown in Eq. (1.3) that the three metric vectors are the three vectors and have demonstrated that there exists a symmetric metric tensor (1.15) and an anti-symmetric metric tensor (1.27). In the complex circular representation the anti-symmetric metric tensor is Eq. (1.28).
Comment Now Evans is using yet another term for the radius-vector of Eq.(1.1) - “displacement vector”. Before, it was a “straight line” and then a “curve”. The term “displacement vector” is certainly better provided we understand that it is a displacement of a point from the origin of the coordinate system! But then there is the term “metric vectors” instead of “vectors of an orthonormal basis”, and “anti-symmetric metric tensor” instead of “two-form”.
To summarize: The first section “Mathematical Prerequisite” of GCUFT by Myron Evans is plagued by elementary mathematical errors and shows that the author has a rather poor understanding of elementary mathematical concepts such as “vector, straight line, metric, differential form, tensor, matrix.” It is also clear that no one has read the text if only to correct evident and elementary calculational mistakes.


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