$S^{ab}$ and $\tilde{S}^{ab}$, manifesting how would matrices
look like if families, suggested by the use of the Clifford algebra in the
{\it spin-charge-family} theory, would be taken into account.

Since one can always embed the spin part in $d=(3+1)$ into $d\ge 5$, and if
at the same time one requires  the Clifford oddness of "basis vectors" in the
whole space $(d-1) +1$ so that spins in higher dimensions manifest as charges
in $d=(3+1)$, the corresponding
creation and annihilation operators fulfill the anticommutation postulates of Dirac.
The single fermion states already anticommute.

We present  matrices for operators $\gamma^a$'s,  $\tilde{\gamma^a}$'s,
$S^{ab}$ and $\tilde{S}^{ab}$ in Sect.~\ref{matrixCliffordDMN}.

Sect.~\ref{evenclifford} demonstrates that  the application of the Clifford odd
operators $\gamma^a$'s and  $\tilde{\gamma^a}$'s, Ref.~[23],
on the Clifford odd  anticommuting "basis vectors" transforms the Clifford
odd "basis vectors" to the Clifford even "basis vector", which commute.

%*TU 29.04.2021\\ 20,21 finish this section!\\

%SHORTEN below!!\\
% *TU READ again and correct if needed
% Let us  add  that,  provided that
% $\omega^{*}_{56 \pm}= \omega_{56 \pm}$ and accepting the expectation values
% $<p_{0\pm}>$ as the "mass" term $m_{\pm}$  (indeed $m^2= m_{+} m_{-}$),
% it then follows $m_{\pm}= -m_{\mp}$, requiring that both, $m_{\pm}$, are imaginary, in
% accordance with Eq.~(\ref{Weylmsolsimple}).

% Possibly also Grassmann?
\subsection{Clifford even creation operators }
\label{evenclifford}

%*TU 22.04.2021 Comment usual Dirac representation!Mention TDN\\.
%The break of symmetries and of mass protection leads to Majorana particles
%Such Majorana particles are the right handed neutrinos together with their
% left handed partners after the break of weak charge symmetry and the
%corresponding mass protection.

In Table~\ref{cliff basis5+1.} the Clifford odd "basis vectors" for $d=(5+1)$ case
are presented as suggested by the {\it spin-charge-family} theory, forming
four families, together with their Hermitian conjugated partners.
Due to the oddness of the "basis vectors" the tensor products, $*_{T}$, of the
"basis vectors" and the basis in the momentum/coordinate space fulfil together
with their corresponding annihilation operators the anticommutation relations
postulated by Dirac, explaining indeed the Dirac's anticommuting postulates.

The remaining $2^{d-1}$ "basis vectors" have an even Clifford character. They are
presented in the lower half part of Table~\ref{Table Clifffourplet.}. The reader can
immediately see that  the multiplication of any  odd "basis vector" by $\gamma^a$
leads to
an even "basis vector". For example, the application of $\gamma^0$ on the first
family member of the first {\it odd I} family generates the first member of the
first {\it even I} family.

All the Clifford even "families" with "family" members of
Table~\ref{Table Clifffourplet.} can be obtained also as algebraic products
($*_{A}$) of the Clifford odd "basis vectors" from the upper half part
of the same table.

All the even families have  one self adjoint "basis vector"; Being a product of
projectors only this "basis vector" is its own Hermitian conjugated partner,
enabling to generate and annihilate the state at the same time, when applying
on the (any one) vacuum state.

Clifford even "basis vectors" commute.

The Clifford even operators $S^{ab}$ transform family members of one
family among themselves, while $\tilde{S}^{ab}$ transform one family
member of any family to the same family member of another family also in the
case of even "basis vectors".

We shall comment these properties in Sect.~\ref{matrixCliffordDMN}.
%*TU Do it!
%In the same table also  superposition of even eigenvectors of the Cartan subalgebras,
%Eq.~(\ref{cartancliff}), are added.

%In the same table also  superposition of even eigenvectors of the Cartan subalgebras,
%Eq.~(\ref{cartancliff}), are added.

%*TU Move Fig. capture down!

% odd I
\begin{table*}
\begin{center}
\begin{tiny}
\begin{minipage}[t]{16.5 cm}
\caption{$2^d=64$ "eigenvectors" of the Cartan
subalgebra, Eq.~(\ref{cartangrasscliff}), of the Clifford  odd and even algebras in
$d=(5+1)$ are presented, divided into four groups, each group with four
"families", each ''family'' with four ''family'' members. Two of four groups are
superposition of products of an odd number of $\gamma^a$'s.  The
"basis vectors", ${\hat b}^{m \dagger}_f$,
Eq.~(\ref{start(2n+1)2cliffgammatilde4n}), in  {\it odd I} group, belong to four
"families" ($f=1 (a),2(b),3(c),4(d)$) with four members ($m=1,2,3,4$), having
their Hermitian conjugated partners, ${\hat b}^{m}_f$, among "basis vectors" of
the {\it odd II} part, denoted with the corresponding "family" and "family"
members ($a_m,b_m,c_m,d_m$) quantum numbers.
The "family" quantum  numbers, determined by
$(\tilde{S}^{03}, \tilde{S}^{12},\tilde{S}^{56})$,
of ${\hat b}^{m \dagger}_f$ are written   above each "family".
The  two groups with the even number of $\gamma^a$'s in a product, {\it even I}
and {\it even II}, have their Hermitian conjugated partners within their own group
each. There are members in each of even groups, which are products of projectors only.
Numbers ---
$\stackrel{03}{\;\,}\;\;\,\stackrel{12}{\;\,}\;\;\,\stackrel{56}{\;\,}$ ---
denote the indexes of the corresponding Cartan subalgebra members
($\tilde{S}^{03}, \tilde{S}^{12}, \tilde{S}^{56}$), Eq.~(\ref{cartangrasscliff}).
In the columns $(7,8,9)$ the eigenvalues of the Cartan subalgebra members
$(S^{03}, S^{12}, S^{56})$, Eq.~(\ref{cartangrasscliff}), are presented. The last
two columns  tell the handedness of "basis vectors" in $d=(5+1)$,
$\Gamma^{(5+1)}$, and of $d=(3+1)$,  $\Gamma^{(3+1)}$, respectively,
defined in Eq.(\ref{hand}).
%Equivalent table for $\tilde{\gamma}^a$'s follow by replacing all $\stackrel{ab}{(k)}$ by
%$\stackrel{ab}{\tilde{(k)}}$ and $\stackrel{ab}{[k]}$ by $\stackrel{ab}{\tilde{[k]}}$.
\label{Table Clifffourplet.}
\end{minipage}
\begin{tabular}{|c|c|c|c|c|c|r|r|r|r|r|}
\hline
$ odd \, I$&$m$&$ f=1 (a)$&$ f=2 (b)$&$ f=3 (c)$&
$ f=4 (d)$&$S^{03}$&$S^{12}$&$S^{56}$&$\Gamma^{(5+1)}$&
\frac{1}{2})$&
\hline
$$&$1$&$\stackrel{03}{(+i)}\stackrel{12}{(+)}\stackrel{56}{(+)}$ &
$\stackrel{03}{[+i]}\stackrel{12}{[+]}\stackrel{56}{(+)}$ &
$\stackrel{03}{[+i]}\stackrel{12}{(+)}\stackrel{56}{[+]}$ &
$$&$2$&    $[-i][-](+) $ & $(-i)(-)(+) $ & $(-i)[-][+] $ & $[-i](-)[+] $ &
$$&$3$&    $[-i] (+)[-]$ & $(-i)[+][-] $ & $(-i)(+)(-) $ & $[-i][+](-) $&
$-\frac{i}{2}$&$\frac{1}{2}$&$-\frac{1}{2}$&$1$&$-1$ \\
$$&$4$&    $(+i)[-][-] $ & $[+i](-)[-] $ & $[+i][-](-) $ & $(+i)(-)(-) $&
$\frac{i}{2}$&$-\frac{1}{2}$&$-\frac{1}{2}$&$1$&$-1$ \\
\hline
$ odd\, II$&$$&$ $&$ $&$ $&
$ $&$S^{03}$&$S^{12}$&$S^{56}$&$\Gamma^{(5+1)}$&
\hline
$$&$$&$(-i)(+)(+)_{d_4}$ & $[-i][+](+)_{d_3}$ & $[-i](+)[+]_{d_2}$ &
$(-i)[+][+]_{d_1}$&$-\frac{i}{2}$&$\frac{1}{2}$&$\frac{1}{2}$&$-1$&$-1$ \\
$$&$$&$[+i][-](+)_{c_4}$ & $(+i)(-)(+)_{c_3}$ & $(+i)[-][+]_{c_2}$ &
$[+i](-)[+]_{c_1}$&$\frac{i}{2}$&$-\frac{1}{2}$&$\frac{1}{2}$&$-1$&$-1$ \\
$$&$$&$[+i](+)[-]_{b_4}$ & $(+i)[+][-]_{b_3}$ & $(+i)(+)(-)_{b_2}$ &
$[+i][+](-)_{b_1}$&$\frac{i}{2}$&$\frac{1}{2}$&$-\frac{1}{2}$&$-1$&$1$ \\
$$&$$&$(-i)[-][-]_{a_4}$ & $[-i](-)[-]_{a_3}$ & $[-i][-](-)_{a_2}$ &
$(-i)(-)(-)_{a_1}$&$-\frac{i}{2}$&$-\frac{1}{2}$&$-\frac{1}{2}$&$-1$&$1$ \\
\hline\hline
$ even\, I$&$m$&$ $&$$&$ $&
$ $&$S^{03}$&$S^{12}$&$S^{56}$&$\Gamma^{(5+1)}$&
\hline
$$&$1$& $[-i](+)(+) $ & $(-i)[+](+) $ & $[-i][+][+] $ & $(-i)(+)[+] $ &
$-\frac{i}{2}$& $\frac{1}{2}$&$\frac{1}{2}$&$-1$&$-1$ \\
$$&$2$&    $(+i)[-](+) $ & $[+i](-)(+) $ & $(+i)(-)[+] $ & $[+i][-][+] $ &
$\frac{i}{2}$&$-\frac{1}{2}$&$\frac{1}{2}$&$-1$&$-1$ \\
$$&$3$&    $(+i)(+)[-] $ & $[+i][+][-] $ & $(+i)[+](-) $ & $[+i](+)(-) $&
$\frac{i}{2}$&$\frac{1}{2}$&$-\frac{1}{2}$&$-1$&$1$ \\
$$&$4$&    $[-i][-][-] $ & $(-i)(-)[-] $ & $[-i](-)(-) $ & $(-i)[-](-) $&
$-\frac{1}{2}$&$-\frac{1}{2}$&$-1$&$1$ \\
\hline
$ even\, II$&$m$&$ $&$ $&$ $&
$ $&$S^{03}$&$S^{12}$&$S^{56}$&$\Gamma^{(5+1)}$&
\hline
$$&$1$&$[+i](+)(+) $ & $(+i)[+](+) $ & $[+i][+][+] $ & $(+i)(+)[+] $ &
$\frac{i}{2}$&$\frac{1}{2}$&$\frac{1}{2}$&$1$&$1$ \\
$$&$2$&$(-i)[-](+) $ & $[-i](-)(+) $ & $(-i)(-)[+] $ & $[-i][-][+] $ &
$-\frac{i}{2}$&$-\frac{1}{2}$&$\frac{1}{2}$&$1$&$1$ \\
$$&$3$&$(-i)(+)[-] $ & $[-i][+][-] $ & $(-i)[+](-) $ & $[-i](+)(-) $&
$-\frac{i}{2}$&$\frac{1}{2}$&$-\frac{1}{2}$&$1$&$-1$ \\
$$&$4$&$[+i][-][-] $ & $(+i)(-)[-] $ & $[+i](-)(-) $ & $(+i)[-](-) $&
$\frac{i}{2}$&$-\frac{1}{2}$&$-\frac{1}{2}$&$1$&$-1$ \\
\hline
\end{tabular}
\end{tiny}
\end{center}
\end{table*}

Let us conclude: The Grassmann algebra  offers the Grassmann odd
creation operators  and their Hermitian conjugated partners annihilation operators,
which anticommute~\footnote{Like in the Clifford case also in the Grassmann
case one can generate creation operators as superposition of tensor products,
$*_{T}$, of the Grassmann odd "basis vectors" and the ordinary, momentum
or coordinate,  basis.},
as well as  Grassmann  even creation operators and their Hermitian conjugated
partners annihilation operators, which commute. The Clifford algebra offers
odd, anticommuting, creation and their Hermitian conjugated partners
annihilation operators, Sect.~\ref{evengrass}, describing the second quantized
fermions with spin, charges and families unified. The commuting Clifford even
creation operators (they have the Hermitian conjugated partners within
the same group)  offer the description of the gauge vector fields to the
corresponding fermion fields.

\subsection{Relations between Clifford algebra and Dirac matrices in the
presence of families in $d=(3+1)$ and $d=(5+1)$}
\label{matrixCliffordDMN}
%\subsection{Relations between Clifford algebra and Dirac matrices in the
%presence of families in $d=(3+1)}$ and $d=(5+1)}$

We learn in Sect.~\ref{internalspace}, in particular after reduction of the
Clifford algebra space, Sect.~\ref{reduction}, and in Sects.~\ref{actionGrassCliff},~
\ref{HilbertCliff0}, that all the fermion states can be represented by the tensor
products, $*_T$, of the finite number of "basis vectors"  and the continuously
infinite basis in the ordinary space, while the total Hilbert space consists of the
tensor products, $*_{T_{H}}$, of all possible numbers of all possible anticommuting
single particle creation operators applying on the vacuum state.
We also learn in Sects.~\ref{TDN1} and \ref{evenclifford} that "basis vectors"
change oddness into evenness if we multiply them by Clifford odd operators
$\gamma^a$'s and $\tilde{\gamma}^a$'s.

The Dirac's single particle states are not Clifford odd and vectors, solving the
Dirac equations, also do not manifest oddness. The fermion states gain the
anticommuting character with the second quantization postulates of Dirac.

In the {\it spin-charge-family} theory fermions already have the anticummuting
nature due to the fact that the theory uses the products of an odd number of
$\gamma^a$'s to describe the internal space of fermions. Correspondingly the
"basis vectors" anticommute, fulfilling the anticommutation relations postulated
by Dirac for the second quantized fermions, and the anticommuting  character
of "basis vectors" transfers also to the states, which are superposition of
tensor products of the "basis vectors" and basis in the momentum or coordinate
space, Eqs.~(\ref{Weylp0}, \ref{Weylp2}),  and are  solutions of equations
of motion.
Let us use the so far several time  discussed toy model in $d=(5+1)$ to
illustrate the application of the operators  $\gamma^a$'s,
$\tilde{\gamma}^a$'s, $S^{ab}$ and $\tilde{S}^{ab}$ on the "basis vectors".

We use in Eq.~(\ref{weylgen0}) the Clifford odd part of "basis vectors" in
$d=(5+1)$, Table~\ref{Table Clifffourplet.}, describing the anticommuting
odd "basis vectors" in $d=(3+1)$ with definite handedness and spin and with
particular charge determined by the spin $S^{56}$, which in this case we neglect.
These states, the solutions of equations of motions for free massless fermions
in $d=(3+1)$, are correspondingly assumed to
have no charge, and are the superposition of the first two members of the third
family and the  first two members of the fourth families, respectively, with the
charge part neglected.

The solutions of equations of motion for free massless particles in $d=(5+1)$ of
Eq.~(\ref{weylgen05+1}) represent fermions and antifermions of the first
family from Table~\ref{cliff basis5+1.} for $p^a=(p^0,  \vec{p},0)$.

The solutions of equations of motion in $d=(13+1)$ for free massless right
handed $u$-quark  of the colour charge $(\frac{1}{2}, \frac{1}{2\sqrt{3}})$
and the "fermion charge" $\tau^4= \frac{1}{6}$ and its left handed antiquark,
Eq.~(\ref{weylgen013+1}), manifest dynamics in $d=(3+1)$ only, if
$p^a=(p^0,  \vec{p},0,0,0,0,\dots,0)$.

In Sect.~\ref{TDN1} the massless and massive solutions in $d=(5+1)$ are

Here we look at the $d=(3+1)$ part of the "basis vectors" belonging to $d=(5+1)$
space, appearing correspondingly in four families. In spaces  with higher
dimensions there would in general case be more families.

The "basis vectors" are arranged as presented in Table~\ref{tableBasVecd4m}
\begin{table}
\begin{tiny}
\begin{center}
\begin{minipage}[t]{16.5 cm}
\caption{In this table $(2^d =)16$ vectors, describing internal space
of fermions in $d=(3+1)$, are presented. Each vector carries the family member quantum
number  $m=(1,2,3,4)$ --- determined by $S^{03}$ and $S^{12}$,
--- and the family quantum number $f=(I,II,III,IV)$ --- determined by
$\tilde{S}^{03}$ and $\tilde{S}^{12}$, Eq.~(\ref{cartangrasscliff}). Vectors
$\psi^m_f$ are obtained by applying $\hat{b}^{m \dagger}_f$ on the vacuum state,
Eq.~(\ref{vac1}, \ref{vac5+1}).
Vectors, that is the family members of any family, split into even (they are sums of
products of an even number of $\gamma^a$'s) and odd (they are sums of products of
an odd number of $\gamma^a$'s). If these vectors are embedded into
$d=(5+1)$ (by being multiplied by an appropriate nilpotent or projector so that they are
of an odd Clifford character), they "gain" charges as presented in Table~\ref{cliff basis5+1.}.}
\label{tableBasVecd4m}
\end{minipage}
\begin{tabular}{|c| c |r r r r r r r r |r r r r r r |}
\hline % $ f {\rm (amily)}\, m$
& $\psi^f_m$ & $\gamma_0\,\psi^f_m$ &
$\gamma_1\,\psi^f_m$ & $\gamma_2\,\psi^f_m$     &$\gamma_3\,\psi^f_m$ &
$\Tilde{\gamma}_0\,\psi^f_m$ &$\Tilde{\gamma}_1\,\psi^f_m$ &
$\Tilde{\gamma}_2\,\psi^f_m$  &$\Tilde{\gamma}_3\,\psi^f_m$ &
$S^{03}$ & $S^{12}$ &$\Tilde{S}^{03}$ &
$\Tilde{S}^{12}$ & $\Gamma^{(3+1)}$ & \vphantom{$b\biggm|b$}
\hline
$\psi^a_1$ & $(+i) (+)$ & $\psi^a_3$ & $\psi^a_4$ & $i\psi^a_4$ & $\psi^a_3$ &
$-i\psi^b_1$ & $-i\psi^c_1$ & $\psi^c_1$ & $-i\psi^b_1$
& $\frac{i}{2}$ & $\frac{1}{2} $ & $\frac{i}{2} $ & $\frac{1}{2} $
& $1 $ & $1 $ \\
$\psi^a_2$ & $[-i] [-]$ & $\psi^a_4$ & $\psi^a_3$ & $-i\psi^a_3$ & $-\psi^a_4$ &
$i\psi^b_2$ & $i\psi^c_2$ & $-\psi^c_2$ & $i\psi^b_2$
& $-\frac{i}{2}$ & $-\frac{1}{2} $ & $\frac{i}{2} $ & $\frac{1}{2} $
& $1 $ & $1 $ \\
$\psi^a_3$ & $[-i] (+)$ & $\psi^a_1$ & $-\psi^a_2$ & $-i\psi^a_2$ & $-\psi^a_1$ &
$i\psi^b_3$ & $i\psi^c_3$ & $-\psi^c_3$ & $i\psi^b_3$
& $-\frac{i}{2}$ & $\frac{1}{2} $ & $\frac{i}{2} $ & $\frac{1}{2} $
& $-1 $ & $1 $ \\
$\psi^a_4$ & $(+i) [-]$ & $\psi^a_2$ & $-\psi^a_1$ & $i\psi^a_1$ & $\psi^a_2$ &
$-i\psi^b_4$ & $-i\psi^c_4$ & $\psi^c_4$ & $-i\psi^b_4$
& $\frac{i}{2}$ & $-\frac{1}{2} $ & $\frac{i}{2} $ & $\frac{1}{2} $
& $-1 $ & $1 $ \\
\hline
$\psi^b_1$ & $[+i] (+)$ & $\psi^b_3$ & $-\psi^b_4$ & $-i\psi^b_4$ & $\psi^b_3$ &
$i\psi^a_1$ & $i\psi^d_1$ & $-\psi^d_1$ & $-i\psi^a_1$
& $\frac{i}{2}$ & $\frac{1}{2} $ & $-\frac{i}{2} $ & $\frac{1}{2} $
& $1 $ & $-1 $ \\
$\psi^b_2$ & $(-i) [-]$ & $\psi^b_4$ & $-\psi^b_3$ & $i\psi^b_3$ & $-\psi^b_4$ &
$-i\psi^a_2$ & $-i\psi^d_2$ & $\psi^d_2$ & $i\psi^a_2$
& $-\frac{i}{2}$ & $-\frac{1}{2} $ & $-\frac{i}{2} $ & $\frac{1}{2} $
& $1 $ & $-1 $ \\
$\psi^b_3$ & $(-i) (+)$ & $\psi^b_1$ & $\psi^b_2$ & $i\psi^b_2$ & $-\psi^b_1$ &
$-i\psi^a_3$ & $-i\psi^d_3$ & $\psi^d_3$ & $i\psi^a_3$
& $-\frac{i}{2}$ & $\frac{1}{2} $ & $-\frac{i}{2} $ & $\frac{1}{2} $
& $-1 $ & $-1 $ \\
$\psi^b_4$ & $[+i] [-]$ & $\psi^b_2$ & $\psi^b_1$ & $-i\psi^b_1$ & $\psi^b_2$ &
$i\psi^a_4$ & $i\psi^d_4$ & $-\psi^d_4$ & $-i\psi^a_4$
& $\frac{i}{2}$ & $-\frac{1}{2} $ & $-\frac{i}{2} $ & $\frac{1}{2} $
& $-1 $ & $-1 $ \\
\hline
$\psi^c_1$ & $(+i) [+]$ & $\psi^c_3$ & $-\psi^c_4$ & $-i\psi^c_4$ & $\psi^c_3$  &
$i\psi^d_1$ & $-i\psi^a_1$ & $-\psi^a_1$ & $i\psi^d_1$
& $\frac{i}{2}$ & $\frac{1}{2} $ & $\frac{i}{2} $ & $-\frac{1}{2} $
& $1 $ & $-1 $ \\
$\psi^c_2$ & $[-i] (-)$ & $\psi^c_4$ & $-\psi^c_3$ & $i\psi^c_3$ & $-\psi^c_4$&
$-i\psi^d_2$ & $i\psi^a_2$ & $\psi^a_2$ & $-i\psi^d_2$
& $-\frac{i}{2}$ & $-\frac{1}{2} $ & $\frac{i}{2} $ & $-\frac{1}{2} $
& $1 $ & $-1 $ \\
$\psi^c_3$ & $[-i] [+]$ & $\psi^c_1$ & $\psi^c_2$ & $i\psi^c_2$ & $-\psi^c_1$ &
$-i\psi^d_3$ & $i\psi^a_3$ & $\psi^a_3$ & $-i\psi^d_3$
& $-\frac{i}{2}$ & $\frac{1}{2} $ & $\frac{i}{2} $ & $-\frac{1}{2} $
& $-1 $ & $-1 $ \\
$\psi^c_4$ & $(+i) (-)$ & $\psi^c_{2}$ & $\psi^c_1$ & $-i\psi^c_1$ &  $\psi^c_{2}$ &
$i\psi^d_4$ & $-i\psi^a_4$ & $-\psi^a_4$ & $i\psi^d_4$
& $\frac{i}{2}$ & $-\frac{1}{2} $ & $\frac{i}{2} $ & $-\frac{1}{2} $
& $-1 $ & $-1 $ \\
\hline
$\psi^d_1$ & $[+i] [+]$ & $\psi^d_3$ & $\psi^d_4$ & $i\psi^d_4$ & $\psi^d_3$&
$-i\psi^c_1$ & $i\psi^b_1$ & $\psi^b_1$ & $i\psi^c_1$
& $\frac{i}{2}$ & $\frac{1}{2} $ & $-\frac{i}{2} $ & $-\frac{1}{2} $
& $1 $ & $1 $ \\
$\psi^d_2$ & $(-i) (-)$ & $\psi^d_4$ & $\psi^d_3$ & $-i\psi^d_3$ & $-\psi^d_4$&
$i\psi^c_2$ & $-i\psi^b_2$ & $-\psi^b_2$ & $-i\psi^c_2$
& $-\frac{i}{2}$ & $-\frac{1}{2} $ & $-\frac{i}{2} $ & $-\frac{1}{2} $
& $1 $ & $1 $ \\
% \hline
$\psi^d_3$ & $(-i) [+]$ & $\psi^d_1$ & $-\psi^d_2$ & $-i\psi^d_2$ & $-\psi^d_1$&
$i\psi^c_3$ & $-i\psi^b_3$ & $-\psi^b_3$ & $-i\psi^c_3$
& $-\frac{i}{2}$ & $\frac{1}{2} $ & $-\frac{i}{2} $ & $-\frac{1}{2} $
& $-1 $ & $1 $ \\
$\psi^d_4$ & $[+i] (-)$ & $\psi^d_2$ & $-\psi^d_1$ & $i\psi^d_1$ & $\psi^d_2$&
$-i\psi^c_4$ & $i\psi^b_4$ & $\psi^b_4$ & $i\psi^c_4$
& $\frac{i}{2}$ & $-\frac{1}{2} $ & $-\frac{i}{2} $ & $-\frac{1}{2} $
& $-1 $ & $1 $ \\
\hline %\hline
\end{tabular}
\end{center}
%Izbrani bazni vektorji za $d=4$.
%%Izbrani bazni vektorji za $d=4$.
\end{tiny}
\end{table}

%As seen in Table~\ref{tableBasVecd4m}
Using the Pauli matrices from Eq.~(\ref{pauli})  as  well as the unit
$4 \times 4$ submatrix and the submatrix
with all the matrix elements equal to zero
\begin{equation}\label{pauli4x4}
\mathbf{1}=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}  \, = \sigma^0\,,
% \end{equation},
\qquad\qquad
%   \begin{equation}\label{dn-mat-mat1}
\mathbf{0}=
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}\,,
\end{equation}
and taking into account Table~\ref{tableBasVecd4m} the matrix representation for
$\gamma^a$'s, $\tilde{\gamma}^a$'s, $S^{ab}$'s and $\tilde{S}^{ab}$'s follow.
% We represent only some of them, the rest can the reader reproduce  in a similar way.
%Make a choice of a few matrices!
\begin{equation}\label{dn-mat-mgamma01}
\gamma^{0} =
\begin{pmatrix}
\begin{smallmatrix}
0 & \sigma^0\\
\sigma^0 & 0
\end{smallmatrix} & \mathbf{0}  & \mathbf{0}  & \mathbf{0} \\
\mathbf{0}  &
\begin{smallmatrix}
0 & \sigma^0\\
\sigma^0 & 0
\end{smallmatrix} & \mathbf{0}  & \mathbf{0} \\
\mathbf{0} & \mathbf{0}  &
\begin{smallmatrix}
0 & \sigma^0\\
\sigma^0 & 0
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
0 & \sigma^0\\
\sigma^0 & 0
\end{smallmatrix}
\end{pmatrix}\,, %%\nonumber\\
% \begin{equation}\label{dn-mat-mgamma1}
\,\gamma^{1} =
\begin{pmatrix}
\begin{smallmatrix}
0 & \sigma^1\\
-\sigma^1 & 0
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
0 & -\sigma^1\\
\sigma^1 & 0
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
0 & -\sigma^1\\
\sigma^1 & 0
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
0 & \sigma^1\\
-\sigma^1 & 0
\end{smallmatrix}
\end{pmatrix}\,, %%\nonumber\\
\end{equation}
\begin{equation}\label{dn-mat-mgamma23}
\gamma^{2} =
\begin{pmatrix}
\begin{smallmatrix}
0 & -\sigma^2\\
\sigma^2 & 0 &
\end{smallmatrix} & 0 & 0 & 0\\
\mathbf{0} &
\begin{smallmatrix}
0 & \sigma^2\\
-\sigma^2 & 0 &
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
0 & \sigma^2\\
-\sigma^2 & 0 &
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
0 & -\sigma^2\\
\sigma^2 & 0 &
\end{smallmatrix}
\end{pmatrix}\,,%%\nonumber\\
%% \end{equation}
\,%% \begin{equation}\label{dn-mat-mgamma3}
\gamma^{3} =
\begin{pmatrix}
\begin{smallmatrix}
0 & \sigma^3\\
-\sigma^3 & 0 &
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
0 & \sigma^3\\
-\sigma^3 & 0 &
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
0 & \sigma^3\\
-\sigma^3 & 0 &
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
0 & \sigma^3\\
-\sigma^3 & 0 &
\end{smallmatrix}
\end{pmatrix}\,,
\end{equation}
manifesting  the $4 \times 4$ substructure (of Dirac matrices) along the diagonal of
$16 \times 16$ matrices.

%  \section{Matrike $\Tilde{\gamma}^a$ za d=4}
% \label{dn-mat-d4tgamma}
The representations of $\tilde{\gamma}^{a}$ do not appear in the Dirac case.
They manifest the off diagonal structure as follows
\begin{equation}\label{dn-mat-mtgamma01}
\Tilde{\gamma}^{0} =
\begin{pmatrix}
\mathbf{0} &
\begin{smallmatrix}
-i\sigma^3 & 0 \\
0 & i\sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0} \\
\begin{smallmatrix}
i\sigma^3 & 0 \\
0 & -i\sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &  \mathbf{0} &
\begin{smallmatrix}
i\sigma^3 & 0 \\
0 & -i\sigma^3
\end{smallmatrix}  \\
\mathbf{0} & \mathbf{0}  &
\begin{smallmatrix}
-i\sigma^3 & 0 \\
0 & i\sigma^3
\end{smallmatrix} & \mathbf{0}
\end{pmatrix}%%\nonumber\\
\,,
%% \begin{equation}\label{dn-mat-mtgamma1}
\, \Tilde{\gamma}^{1} =
\begin{pmatrix}
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
-i\sigma^3 & 0 \\
0 & i\sigma^3
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
i\sigma^3 & 0 \\
0 & -i\sigma^3
\end{smallmatrix} \\
\begin{smallmatrix}
-i\sigma^3 & 0 \\
0 & i\sigma^3
\end{smallmatrix} & \mathbf{0}  & \mathbf{0} & \mathbf{0} \\
\mathbf{0} &
\begin{smallmatrix}
i\sigma^3 & 0 \\
0 & -i\sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0}
\end{pmatrix}\,, %%\nonumber\\
\end{equation}
\begin{equation}\label{dn-mat-mtgamma23}
\Tilde{\gamma}^{2} =
\begin{pmatrix}
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & -\sigma^3
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
-\sigma^3 & 0 \\
0 & \sigma^3
\end{smallmatrix} \\
\begin{smallmatrix}
-\sigma^3 & 0 \\
0 & \sigma^3
\end{smallmatrix} & \mathbf{0}  & \mathbf{0} & \mathbf{0} \\
\mathbf{0} &
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & -\sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0}
\end{pmatrix}\,, %%\nonumber\\
%% \end{equation}
%% \begin{equation}\label{dn-mat-mtgamma3}
\, \Tilde{\gamma}^{3} =
\begin{pmatrix}
\mathbf{0} &
\begin{smallmatrix}
-i\sigma^3 & 0 \\
0 & i\sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0} \\
\begin{smallmatrix}
-i\sigma^3 & 0 \\
0 & i\sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &  \mathbf{0} &
\begin{smallmatrix}
-i\sigma^3 & 0 \\
0 & i\sigma^3
\end{smallmatrix}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
-i\sigma^3 & 0 \\
0 & i\sigma^3
\end{smallmatrix} & \mathbf{0}
\end{pmatrix}\,.
\end{equation}

%\section{Matrike $S^{ab}$ za d=4}
Matrices $S^{ab}$ have again along the diagonal the $4 \times 4$ substructure,
repeating, up to a phase, as expected, the corresponding Dirac matrices,
%. They manifest the repetition of the Dirac $4 \times 4$ matrices, up to a phase,
since the Dirac $S^{ab}$ do not distinguish among families.
\begin{equation}\label{dn-mat-mS0102}
S^{01} = \frac{i}{2}
\begin{pmatrix}
\begin{smallmatrix}
\sigma^1 & 0 \\
0 & -\sigma^1
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
-\sigma^1 & 0 \\
0 & \sigma^1
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
-\sigma^1 & 0 \\
0 & \sigma^1
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^1 & 0 \\
0 & -\sigma^1
\end{smallmatrix}
\end{pmatrix}\,, %%\nonumber\\
%% \end{equation}
%% \begin{equation}\label{dn-mat-mS02}
\, S^{02} = \frac{i}{2}
\begin{pmatrix}
\begin{smallmatrix}
-\sigma^2 & 0 \\
0 & \sigma^2
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
\sigma^2 & 0 \\
0 & -\sigma^2
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^2 & 0 \\
0 & -\sigma^2
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
-\sigma^2 & 0 \\
0 & \sigma^2
\end{smallmatrix}
\end{pmatrix}\,,  %%\nonumber\\
\end{equation}
\begin{equation}\label{dn-mat-mS0312}
S^{03} = \frac{i}{2}
\begin{pmatrix}
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & -\sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & -\sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & -\sigma^3
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & -\sigma^3
\end{smallmatrix}
\end{pmatrix}\,,  %%\nonumber\\
%% \end{equation}
%% \begin{equation}\label{dn-mat-mS12}
\, S^{12} = \frac{1}{2}
\begin{pmatrix}
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & \sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & \sigma^3
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & \sigma^3
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^3 & 0 \\
0 & \sigma^3
\end{smallmatrix}
\end{pmatrix}\,, %%\nonumber\\
\end{equation}
\begin{equation}\label{dn-mat-mS1323}
S^{13} = \frac{1}{2}
\begin{pmatrix}
\begin{smallmatrix}
\sigma^2 & 0 \\
0 & \sigma^2
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
-\sigma^2 & 0 \\
0 & -\sigma^2
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
-\sigma^2 & 0 \\
0 & -\sigma^2
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^2 & 0 \\
0 & \sigma^2
\end{smallmatrix}
\end{pmatrix}\,, %%\nonumber\\
%% \end{equation}
%% \begin{equation}\label{dn-mat-mS23}
S^{23} = \frac{1}{2}
\begin{pmatrix}
\begin{smallmatrix}
\sigma^1 & 0 \\
0 & \sigma^1
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
-\sigma^1 & 0 \\
0 & -\sigma^1
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
-\sigma^1 & 0 \\
0 & -\sigma^1
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
\sigma^1 & 0 \\
0 & \sigma^1
\end{smallmatrix}
\end{pmatrix}\,.
\end{equation}
\begin{equation}\label{dn-mat-mGamma31}
\Gamma^{(3+1)}=-4i S^{03}S^{12}=
\begin{pmatrix}
\begin{smallmatrix}
1 & 0 \\
0 & -1
\end{smallmatrix} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\begin{smallmatrix}
1 & 0 \\
0 & -1
\end{smallmatrix} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
1 & 0 \\
0 & -1
\end{smallmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\begin{smallmatrix}
1 & 0 \\
0 & -1
\end{smallmatrix}
\end{pmatrix}\,.
\end{equation}

% \section{Matrike $\Tilde{S}^{ab}$ za d=4}
The matrices of the operators $\tilde{S}^{ab}$ have again off diagonal
$4 \times 4$ substructure, except
$\tilde{S}^{03}$ and $\tilde{S}^{12}$, which~are~diagonal.
\begin{equation}\label{dn-mat-mtS010203}
\Tilde{S}^{01} = -\frac{i}{2}
\begin{pmatrix}
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\mathbf{1}\\
\mathbf{0} & \mathbf{0} &
\mathbf{1} & \mathbf{0} \\
\mathbf{0} &
\mathbf{1} & \mathbf{0}  & \mathbf{0} \\
\mathbf{1} & \mathbf{0} & \mathbf{0} & \mathbf{0}
\end{pmatrix}\,, %%\nonumber\\
%% \end{equation}
%% \begin{equation}\label{dn-mat-mtS02}
\, \Tilde{S}^{02} = \frac{1}{2}
\begin{pmatrix}
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\mathbf{1}\\
\mathbf{0} & \mathbf{0} &
\mathbf{1} & \mathbf{0} \\
\mathbf{0} &
-\mathbf{1} & \mathbf{0}  & \mathbf{0} \\
-\mathbf{1} & \mathbf{0} & \mathbf{0} & \mathbf{0}
\end{pmatrix}\,,  %%\nonumber\\
%% \end{equation}
%% \begin{equation}\label{dn-mat-mtS0312}
\, \Tilde{S}^{03} = \frac{i}{2}
\begin{pmatrix}
\mathbf{1} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
-\mathbf{1} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
\mathbf{1} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\end{pmatrix}\,,  %%\nonumber\\
\end{equation}
\begin{equation}\label{dn-mat-mtS121323}
\, \Tilde{S}^{12} = \frac{1}{2}
\begin{pmatrix}
\mathbf{1} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
\mathbf{1}  & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
-\mathbf{1}  & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\end{pmatrix}\,,  %%\nonumber\\
\, \Tilde{S}^{13} = \frac{i}{2}
\begin{pmatrix}
\mathbf{0} & \mathbf{0} & \mathbf{0} &
-\mathbf{1}  \\
\mathbf{0} & \mathbf{0} &
\mathbf{1} & \mathbf{0} \\
\mathbf{0} &
-\mathbf{1} & \mathbf{0}  & \mathbf{0} \\
\mathbf{1} & \mathbf{0} & \mathbf{0} & \mathbf{0}
\end{pmatrix}\,, %%\nonumber\\
%% \end{equation}
%% \begin{equation}\label{dn-mat-mtS23}
\, \Tilde{S}^{23} = \frac{1}{2}
\begin{pmatrix}
\mathbf{0} & \mathbf{0} & \mathbf{0} &
-\mathbf{1}  \\
\mathbf{0} & \mathbf{0} &
\mathbf{1} & \mathbf{0} \\
\mathbf{0} &
\mathbf{1} & \mathbf{0}  & \mathbf{0} \\
-\mathbf{1} & \mathbf{0} & \mathbf{0} & \mathbf{0}
\end{pmatrix}\,.
\end{equation}

\begin{equation}\label{dn-mat-mtGamma31}
\Tilde{\Gamma}^{(3+1)}=-4i \Tilde{S}^{03}\Tilde{S}^{12}=
\begin{pmatrix}
\mathbf{1} & \mathbf{0} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} &
-\mathbf{1} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} &
-\mathbf{1} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \mathbf{0} &
\mathbf{1}
\end{pmatrix}\,.
\end{equation}

Let us now conclude this appendix:
We start with the operators $\gamma^a$'s and $\tilde{\gamma}^a$'s and
express the  "basis vectors" as  superposition of and an odd number of
Here  we present the matrix representation of the operators $\gamma^a$'s,
$\tilde{\gamma}^a$'s, $S^{ab}$'s and $\tilde{S}^{ab}$'s, operating
on the "basis vectors", which are eigenvectors of the Cartan subalgebra of the
Lorentz algebra, $S^{ab}$'s and $\tilde{S}^{ab}$'s. This matrices can be
compared with the Pauli matrices and
$\gamma^m$'s and $S^{mn}$'s matrices of Dirac, after we neglect the charge
part in the $5^{th}$ and the $6^{th}$ dimension of the internal space of fermions.

Since there are several families the representations of the operators
$\gamma^a$'s, $\tilde{\gamma}^a$'s, $S^{ab}$'s and $\tilde{S}^{ab}$'s
manifest the existence of families.

The states on which the Pauli matrices  and the Dirac's $\gamma^a$ matrices apply
are just columns of numbers, with no anticommuting character, which "basis vectors"
do have.

% *TU this should come to the conclusions
The  Clifford algebra enables  to create creation operators which define, applying
on the vacuum state, the single particle states with the odd Clifford character.
Since the oddness of (the finite number of) "basis vectors" transfers in the
tensor products, $*_{T}$,  with the continuously infinite basis in ordinary space
to the creation and annihilation operators, making them Clifford odd, the Dirac
postulates are not needed.
The use of the Clifford algebra for the description of the internal space explains the
Dirac's second quantized postulates.

In Sect.~\ref{creationannihilationtensor} we compare the way of the second
quantization of fermions in the {\i spin-charge-family} theory with the Dirac's one.

% Table~\ref{Table Clifffourplet.} represents besides Clifford odd anticommuting
% "basis vectors",
% also Clifford even commuting "basis vectors" to point out that application of
% $\gamma^a$'s on the Clifford odd creation operators has no meaning.

%*TU ADD more.

\section{ Understanding  triangle anomalies cancellation in standard model}
\label{appanomalies}

%*TU Shorten and Correct

We clarify here the difference in explaining  the "miraculous" anomalies cancellation
in the {\it standard model} in the $SO(10)$ unifying theories and the
{\it spin-charge-family} theory. Since the $SO(10)$ group is subgroup of
$SO(13,1)$ group, the difference occurs from the fact that in  the
{\it spin-charge-family} theory the spins and charges are unified, while in the $SO(10)$
theories they are not. One sees in Table ~\ref{Table so13+1.} that handedness
and charges of quarks and leptons and of antiquarks and antileptons are uniquely
determined --- fermions and anti-fermions belong to the same irreducible representation
of the Lorentz group, while in the unifying theories handedness and charges must be
correlated "by hand".
We follow here mainly Ref.~[20].

In $d=(2n)$-dimensional space-time massless fermions contribute through the
one-loop (n + 1)-angle diagram in general an anomalous (infinite) function, which
causes the current non-conservation and contributes to the gauge non-invariance
of the action~[56, 57].

To the  triangle anomalies the right-handed spinors (fermions) and the right handed
anti-spinors (anti-fermions) contribute with the opposite sign than the left handed
spinors and the left handed anti-spinors. Their common contribution to anomalies is
proportional to~[56]
\begin{eqnarray}
\label{anomaly0}
(\sum_{(A,i,B,j,C,k)_{L \,\bar{L}}} Tr [\tau^{Ai} \, \tau^{Bj} \,\tau^{Ck}] -
\sum_{(A,i,B,j,C,k)_{R \,\bar{R}}} Tr [\tau^{Ai} \, \tau^{Bj} \,\tau^{Ck}]\,) \,,
\end{eqnarray}
where $\tau^{Ai}$ are in the {\it standard model} the generators of the infinitesimal
transformation of the groups $SU(3), SU(2)$ and $U(1)$, while in the
{\it spin-charge-family} theory $\tau^{Ai}$ are the infinitesimal generators of the
irreducible subgroups of the starting orthogonal group $SO(2(2n+1)-1,1)$, $n=3$
(which include all the {\it standard model} groups, offering correspondingly the
explanation for their origin), Eqs.~(\ref{so1+3}, \ref{so42}, \ref{so64}).
The traces run over the representations of one massless family of the left handed
fermions and anti-fermions, denoted by ${}_{L \,\bar{L}}$, and the right handed
fermions and anti-fermions, denoted by ${}_{R \,\bar{R}}$.

Embedding the {\it standard model} groups into the orthogonal group $SO(13,1)$
explains elegantly the "miraculous" cancellation of the triangle anomalies in the
{\it standard model}. Embedding the  {\it standard model} groups into the
orthogonal group $SO(10)$ group explains the "miraculous" cancellation of the
triangle anomalies if one correlates charges and handedness.

Table~\ref{Table so13+1.}  presents spinor handedness ($\Gamma^{(3,1)}$),
their spin ($S^{12}$), weak charge $(\tau^{13})$, the second $SU(2)_{II}$
$(\tau^{23})$ charge (arising together with $SU(2)_{I}$ from $SO(4)$), their
colour charge $(\tau^{33},\tau^{38})$  (arising together with $U(1)_{II}$ from
$SO(6)$), and the "fermion charge" ($\tau^{4}$, the generator of $U(1)_{II}$).
The hyper charge is $Y=(\tau^{23} +\tau^{4})$,  the electromagnetic charge is
$Q=(\tau^{13}+ Y)$.

%The cancellation of the triangle anomalies of the massless quarks and leptons and their anti-particles
% looks from the point of view of the {\it standard model} "miraculous".

The triangle anomaly  of the {\it standard model}  occurs if the traces in
Eq.(\ref{anomaly0}) are not zero for either the left handed quarks and leptons and
anti-quarks and anti-leptons or the right handed quarks and leptons and anti-quarks
and anti-leptons for the Feynman triangle diagrams in which the
gauge vector fields of the type
\begin{eqnarray}
\label{anomalyparticular}
&& U(1)\times U(1) \times U(1)\, ,\nonumber\\
&& SU(2)\times SU(2) \times U(1)\, ,\nonumber\\
&& SU(3)\times SU(3) \times SU(3)\, ,\nonumber\\
&& SU(3)\times SU(3) \times U(1)\, ,\nonumber\\
&& U(1) \times {\rm gravitational}\,
\end{eqnarray}
contribute to the triangle anomaly.

%*TU 26.04.2021

Table~\ref{Table SMandSCFTspinors.} presents in the first seven columns (up to $||$)
by the {\it standard model} assumed properties of the members of any massless
family, running in the triangle. The last two columns, taken from
Table~\ref{Table so13+1.}, describe additional properties which quarks and leptons and
anti-quarks and anti-leptons would have, if the {\it standard model} groups
$SO(3,1)$, $SU(2), SU(3)$ and $U(1)$ are embedded into the $SO(13,1)$ group.
As already pointed out, the $SO(10)$ unifying theories do from the
point of the charges  the same, but do  correlate handedness and charges in the
same way as in the {\t standard model}.
We  comment these last two columns in particular in Sect.~\ref{fermionsbosonslowE}.

In the {\it spin-charge-family} theory the family quantum numbers are determined by
the second kind of the Clifford algebra objects $\tilde{S}^{ab}$, which commute
with $S^{ab}$ describing spins and charges. Correspondingly the spins and charges
are the same for all the families.
\begin{table}
\begin{center}
\begin{minipage}[t]{16.5 cm}
\caption{Properties  of the left handed quarks and leptons and
of the left handed anti-quarks and anti-leptons (the first $16$ lines)  and of the right
handed quarks and leptons and the right handed anti-quarks and anti-leptons (the
second $16$ lines), as assumed by the {\it standard model}, are presented in the
first eight columns. In the last two columns the two quantum numbers are added,
which fermions and anti-fermions would have if the {\it standard model} groups
$SO(3,1)$, $SU(2)$, $SU(3)$ and $U(1)$ are embedded into the $SO(13,1)$ group.
All the infinitesimal generators of the subgroups of the orthogonal group $SO(13,1)$,
representing charges in $d=(3+1)$, are in the {\it spin-charge-family} theory the
superposition of the generators $S^{st}$,
Eqs.~(\ref{so1+3}, \ref{so42}, \ref{so64}).
The handedness is defined in Eq.(\ref{hand}).
The whole quark part appears, due to the colour charges, three times.
These quantum numbers are the same for all the families.}
\label{Table SMandSCFTspinors.}
\end{minipage}
\begin{tabular}{|r r | c r r r r r ||r r|}
\hline
& & hand-   & weak    & hyper  & colour&charge & elm   & $SU(2)_{II}$&$U(1)_{II}$\\
&&edness  & charge  & charge &        &          &charge&     charge     &  charge     \\
$i_{L}$&name    &$ \Gamma^{(3,1)}$&$ \tau^{13}$  & $ Y$ & $\tau^{33}$&$\tau^{38}$
\hline
$1_{L}$&$ u_{L} $&$ -1     $&$\frac{1}{2}$ &$ \frac{1}{6}$&$\frac{1}{2} $&$\frac{1}{2\sqrt{3}}$&
$ \frac{2}{3}$&0&$\frac{1}{6}$\\
$2_{L}$&$ d_{L} $&$ -1    $&$ -\frac{1}{2}$ &$ \frac{1}{6}$&$\frac{1}{2}$&$\frac{1}{2\sqrt{3}}$ &
$3_{L}$&$ u_{L} $&$ -1     $&$\frac{1}{2}$ &$ \frac{1}{6}$&$-\frac{1}{2} $&$\frac{1}{2\sqrt{3}}$&
$ \frac{2}{3}$&0&$\frac{1}{6}$\\
$4_{L}$&$ d_{L} $&$ -1    $&$ -\frac{1}{2}$ &$ \frac{1}{6}$&$-\frac{1}{2}$&$\frac{1}{2\sqrt{3}}$ &
$5_{L}$&$ u_{L} $&$ -1     $&$\frac{1}{2}$ &$ \frac{1}{6}$&$ 0 $&$-\frac{1}{\sqrt{3}}$&
$ \frac{2}{3}$&0&$\frac{1}{6}$\\
$6_{L}$&$ d_{L} $&$ -1    $&$ -\frac{1}{2}$ &$ \frac{1}{6}$&$ 0$&$-\frac{1}{\sqrt{3}}$ &
\hline
$7_{L}$&$\nu_{L} $&$ -1  $&$ \frac{1}{2}$&$ -\frac{1}{2}$& 0&0&  0  &$0$&$ -\frac{1}{2}$  \\
$8_{L}$&$ e^{L}  $&$ -1  $&$-\frac{1}{2}$&$ -\frac{1}{2}$& 0&0&$-1$&$0$&$ -\frac{1}{2}$ \\
\hline \hline
$9_{L}$&$ \bar{u}{L} $&$ -1$&$  0 $ &$ -\frac{2}{3}$& $ - \frac{1}{2}$ &$-\frac{1}{2\sqrt{3}}$
&$- \frac{2}{3}$ &$ - \frac{1}{2}$ &$ - \frac{1}{6}$\\
$10_{L}$&$ \bar{d}{L} $&$ -1$&$  0 $ &$  \frac{1}{3}$& $ - \frac{1}{2}$ &$-\frac{1}{2\sqrt{3}}$
&$  \frac{1}{3}$ &$   \frac{1}{2}$ &$ - \frac{1}{6}$\\
$11_{L}$&$ \bar{u}{L} $&$ -1$&$  0 $ &$ -\frac{2}{3}$& $  \frac{1}{2}$ &$-\frac{1}{2\sqrt{3}}$
&$- \frac{2}{3}$ &$ - \frac{1}{2}$ &$ - \frac{1}{6}$\\
$12_{L}$&$ \bar{d}{L} $&$ -1$&$  0 $ &$  \frac{1}{3}$& $  \frac{1}{2}$ &$-\frac{1}{2\sqrt{3}}$
&$  \frac{1}{3}$ &$   \frac{1}{2}$ &$ - \frac{1}{6}$\\
$13_{L}$&$ \bar{u}{L} $&$ -1$&$  0 $ &$ -\frac{2}{3}$& $ 0 $&$ \frac{1}{\sqrt{3}}$
&$- \frac{2}{3}$ &$ - \frac{1}{2}$ &$ - \frac{1}{6}$\\
$14_{L}$&$ \bar{d}{L} $&$ -1$&$  0 $ &$  \frac{1}{3}$& $ 0 $&$\frac{1}{\sqrt{3}}$
&$  \frac{1}{3}$ &$   \frac{1}{2}$ &$ - \frac{1}{6}$\\
\hline
$15_{L}$&$\bar{\nu}_{L}$&$ -1$&$ 0 $&$ 0 $& 0&0&$0$&$- \frac{1}{2}$&$ \frac{1}{2}$  \\
$16_{L}$&$\bar{e}_{L}   $&$ -1$&$ 0 $&$ 1 $& 0&0&$1$&$  \frac{1}{2}$&$ \frac{1}{2}$  \\
\hline \hline\hline
$1_{R}$&$u_{R}   $&$ 1$& $0$ & $  \frac{2}{3}$&$ \frac{1}{2}$ &$\frac{1}{2\sqrt{3}}$
&$ \frac{ 2}{3}$&$ \frac{1}{2}$&$ \frac{1}{6}$ \\
$2_{R}$&$d_{R}   $&$ 1$& $0$ & $ -\frac{1}{3}$&$ \frac{1}{2}$ &$\frac{1}{2\sqrt{3}}$
&$-\frac{1}{3}$&$- \frac{1}{2}$&$ \frac{1}{6}$  \\
$3_{R}$&$u_{R}   $&$ 1$& $0$ & $  \frac{2}{3}$&$- \frac{1}{2}$ &$\frac{1}{2\sqrt{3}}$
&$ \frac{ 2}{3}$&$ \frac{1}{2}$&$ \frac{1}{6}$ \\
$4_{R}$&$d_{R}   $&$ 1$& $0$ & $ -\frac{1}{3}$&$- \frac{1}{2}$ &$\frac{1}{2\sqrt{3}}$
&$-\frac{1}{3}$&$- \frac{1}{2}$&$ \frac{1}{6}$  \\
$5_{R}$&$u_{R}   $&$ 1$& $0$ & $  \frac{2}{3}$&$ 0  $ &$-\frac{1}{\sqrt{3}}$
&$ \frac{ 2}{3}$&$ \frac{1}{2}$&$ \frac{1}{6}$ \\
$6_{R}$&$d_{R}   $&$ 1$& $0$ & $ -\frac{1}{3}$&$ 0  $ &$-\frac{1}{\sqrt{3}}$
&$-\frac{1}{3}$&$- \frac{1}{2}$&$ \frac{1}{6}$  \\
\hline
$7_{R}$&$\nu_{R}$&$ 1$& $0$  & $ 0$&0&0&   0 &$  \frac{1}{2}$&$- \frac{1}{2}$           \\
$8_{R}$&$ e_{R}  $&$ 1$& $0$  & $-1$&0&0&$-1$&$- \frac{1}{2}$&$- \frac{1}{2}$  \\
\hline\hline
%anti-particles%%%%12.08.2016 8:43%%%%12.08.2016 8:43
$9_{R}$&$\bar{u}_{R}   $&$ 1$& $ -\frac{1}{2}$ & $ - \frac{1}{6}$&$ -\frac{1}{2}$ &$-\frac{1}{2\sqrt{3}}$
&$ - \frac{ 2}{3}$&0& $ - \frac{1}{6}$\\
$10_{R}$&$\bar{d}_{R}   $&$ 1$&$  \frac{1}{2}$ & $  - \frac{1}{6}$&$ -\frac{1}{2}$ &$-\frac{1}{2\sqrt{3}}$
&$    \frac{1}{3}$&0& $ - \frac{1}{6}$\\
$11_{R}$&$\bar{u}_{R}   $&$ 1$& $ -\frac{1}{2}$ & $ -  \frac{1}{6}$&$ \frac{1}{2}$ &$-\frac{1}{2\sqrt{3}}$
&$ - \frac{ 2}{3}$&0& $ - \frac{1}{6}$\\
$12_{R}$&$\bar{d}_{R}   $&$ 1$&$  \frac{1}{2}$ & $ -  \frac{1}{6}$&$  \frac{1}{2}$ &$-\frac{1}{2\sqrt{3}}$
&$    \frac{1}{3}$&0& $ - \frac{1}{6}$\\
$13_{R}$&$\bar{u}_{R}   $&$ 1$& $ -\frac{1}{2}$ & $ -  \frac{1}{6}$&$ 0 $ &$ \frac{1}{\sqrt{3}}$
&$ - \frac{ 2}{3}$&0& $ - \frac{1}{6}$\\
$14_{R}$&$\bar{d}_{R}   $&$ 1$&$  \frac{1}{2}$ & $  -  \frac{1}{6}$&$ 0 $ &$ \frac{1}{\sqrt{3}}$
&$    \frac{1}{3}$&0& $ - \frac{1}{6}$\\
\hline
$15_{R}$&$\bar{\nu}_{R}$&$ 1$& $ -\frac{1}{2}$ & $  \frac{1}{2}$&0&0&$0$&0&$\frac{1}{2}$        \\
$16_{R}$&$\bar{e}_{R}   $&$ 1$& $  \frac{1}{2}$ & $  \frac{1}{2}$&0&0&$1$&0&$\frac{1}{2}$ \\
\hline\hline
\end{tabular}
\end{center}
%\caption{\label{Table I.}}
% \end{table}
\end{table}

To calculate the traces required in Eq.~(\ref{anomaly0}) for the triangle anomalies of
Eq.~(\ref{anomalyparticular}) the quantum numbers of the left handed spinors and
anti-spinors, as well as of the right handed spinors and anti-spinors, presented in
Table~\ref{Table SMandSCFTspinors.},
are needed.

Let us calculate the traces, Eq.~(\ref{anomalyparticular}), for possible anomalous
triangle diagrams in $d=(3+1)$. One must evaluate the trace of the product  of
three generators and sum the trace over all the states of either the left handed
members ---  $16$ states presented in the first part of
Table~\ref{Table SMandSCFTspinors.} --- or the right handed members --- $16$
states presented in the second part of Table~\ref{Table SMandSCFTspinors.}.
Let us recognize again that in the case of embedding the {\it standard model}
groups into $SO(13,1)$ we have $Y=(\tau^{4} + \tau^{23}) $.

For the triangle Feynman diagram,  to which three hyper $U(1)$ boson fields
contribute, we must evaluate  $\sum_{i} Tr (Y_{i})^3$, in which the sum runs
over all the members ($i=(1,..,16)$) of the left handed spinors and anti-spinors,
and of the right handed spinors and anti-spinors separately.
In the case of embedding the {\it standard model} groups into $SO(13,1)$ we have
\begin{eqnarray}
\label{3Ytriangle}
\sum_{i_{L,R}}\, (Y_{i_{L,R}})^3&=&\sum_{i_{L,R}}\,
(\tau^{4}_{i_{L,R}} + \tau^{23}_{i_{L,R}})^{3}
\nonumber\\
&=&\sum_{i_{L,R}}\, (\tau^{4}_{i_{L,R}})^{3} + \sum_{i_{L,R}}\, (\tau^{23}_{i_{L,R}})^{3}
\nonumber\\
&+&\sum_{i_{L,R}}\, 3 \cdot (\tau^{4}_{i_{L,R}})^{2} \cdot \tau^{23}_{i_{L,R}}
+  \sum_{i_{L,R}}\, 3 \cdot \tau^{4}_{i_{L,R}} \cdot  (\tau^{23}_{i_{L,R}})^{2}\,,
\end{eqnarray}
for either the left, $i_{L}$, or the right, $i_{R}$, handed members.
Table~\ref{Table SMandSCFTspinors.}  demonstrates  clearly that
$ (Y_{i_{L,R}})^3=0$ without really making any algebraic evaluation.
Namely, the last column of Table~\ref{Table SMandSCFTspinors.}
manifests that $ \sum_{i_{L}}\,
(\tau^{4}_{i_{L}})^{3} =0 $ [in details: $\sum_{i_{L}}\, (\tau^{4}_{i_{L}})^{3}  $
$ = 2 \cdot 3 \cdot (\frac{1}{6})^{3} + 2 \cdot 3 \cdot (-\frac{1}{6})^{3} +
2 \cdot (-\frac{1}{2})^{3} + 2 \cdot (\frac{1}{2})^{3}=0  $].
%$=\sum_{i_{R}}\,  (\tau^{4}_{i_{R}})^{3} =0$].
Table~\ref{Table SMandSCFTspinors.}  also demonstrates  (the last but one column)
that $ \sum_{i_{L}}\, (\tau^{23}_{i_{L}})^{3} =0 $
$[=(3+1) \cdot ((-\frac{1}{2})^{3}  +  (\frac{1}{2})^{3}))]$, and that also
$ \sum_{i_{R}}\,  (\tau^{23}_{i_{R}})^{3} =0 $
$[= (3+1) \cdot ((\frac{1}{2})^{3} +
(- \frac{1}{2})^{3})]$.

From Table~\ref{Table SMandSCFTspinors.}  one sees also (without calculating) that
$\sum_{i_{L}}\, 3 \cdot (\tau^{4}_{i_{L}})^{2} \cdot \tau^{23}_{i_{L}} =0$,
in particular
$\sum_{i_{L}}\, 3 \cdot (\tau^{4}_{i_{L}})^{2} \cdot \tau^{23}_{i_{L}} $
$=3.\{ ((\frac{1}{2})^2 \cdot (-\frac{1}{2} +\frac{1}{2})+ 3 \cdot (-\frac{1}{6})^2
\cdot (-\frac{1}{2} +\frac{1}{2})\}$, as well as  that $\sum_{i_{R}}\, 3 \cdot
(\tau^{4}_{i_{R}})^{2} \cdot \tau^{23}_{i_{R}} =0$ [$=3\cdot\{ ((-\frac{1}{2})^2
\cdot (\frac{1}{2} +
(-\frac{1}{2}))+ 3 \cdot (\frac{1}{6})^2  \cdot (\frac{1}{2} + (-\frac{1}{2}))\}$].

That the last term in Eq.~(\ref{3Ytriangle}) is zero for either the left or the
right handed spinors can also easily be seen just by looking at
Table~\ref{Table SMandSCFTspinors.} [or in details:
$ \sum_{i_{L}}\, 3 \cdot \tau^{4}_{i_{L}} \cdot  (\tau^{23}_{i_{L}})^{2}=0$
$=3\cdot \{(\frac{1}{2} ((\frac{1}{2})^2 +(-\frac{1}{2})^2) + 3\cdot (-\frac{1}{6})
((\frac{1}{2})^2 +(-\frac{1}{2})^2))\} $, as well as that
$ \sum_{i_{R}}\, 3 \cdot \tau^{4}_{i_{L}} \cdot  (\tau^{23}_{i_{R}})^{2}=0$
$=3\cdot \{(-\frac{1}{2} ((\frac{1}{2})^2 +(-\frac{1}{2})^2) + 3\cdot (\frac{1}{6})
((\frac{1}{2})^2 +(-\frac{1}{2})^2))\} $].

Since all the members belong to one spinor representation, it is straightforwardly
that all the triangle traces are zero, if the {\it standard model}  groups are the
subgroups of the orthogonal group  $SO(13,1)$.

There is no need for a detailed calculations, since a look in
Table~\ref{Table SMandSCFTspinors.} gives immediately the answer.

From only the {\it standard model} assumptions point of view the cancellation of
the triangle anomalies does look  miraculous. For our $\sum_{i_{L,R}}\, (Y_{i_{L,R}})^3$
one obtains for the left handed members:
[$3 \cdot2 \cdot (\frac{1}{6})^3 + 2 \cdot (-\frac{1}{2})^3 +
3 \cdot  ((-\frac{2}{3})^3 + (\frac{1}{3})^3) + 1^3)$ ], and for the right handed
members:  [$3 \cdot ((\frac{2}{3})^3 + (-\frac{1}{3})^3)+
(- 1)^3) +3 \cdot 2 \cdot (-\frac{1}{6})^3 + 2 \cdot (\frac{1}{2})^3 $ ].

\section{Norms in Grassmann space and Clifford space}
\label{normgrass}

% Change below!!

Let us define the integral over the Grassmann  space~[9] of two functions of the
Grassmann coordinates $<{\cal {\bf B}}|\theta> <{\cal {\bf C}}|\theta>$,  $<{\cal {\bf B}}| \theta>=
<\theta | {\cal {\bf B}}>^{\dagger}$,
\[<{\cal {\bf B}}|\theta>= \sum_{k=0}^{d} b_{a_1\dots a_k}
\theta^{a_1}\cdots \theta^{a_k},\]
by requiring
\begin{eqnarray}
\label{grassintegral}
&&\{ d\theta^a, \theta^b \}_{+} =0\,, \,\;\;  \int d\theta^a  =0\,,\,\;\;
\int d\theta^a \theta^a =1\,,\;\; \nonumber\\
&&\int d^d \theta \,\,\theta^0 \theta^1 \cdots \theta^d =1\,,
\nonumber\\
&&d^d \theta =d \theta^d \dots d\theta^0\,,\,\;\; %\nonumber\\
\omega = \prod^{d}_{k=0}(\frac{\partial}{\;\,\partial \theta_k} + \theta^{k})\,,
\end{eqnarray}
with $ \frac{\partial}{\;\,\partial \theta_a} \theta^c = \eta^{ac}$. We shall use the weight function~%
[9]
$\omega= \prod^{d}_{k=0}(\frac{\partial}{\;\,\partial \theta_k} + \theta^{k})$ to define the scalar
product  in Grassmann space $<{\cal {\bf B}}|{\cal {\bf C}} >$
\begin{eqnarray}
\label{grassnorm}
<{\cal {\bf B}}|{\cal {\bf C}} > &=&  \int %d^{d-1} x
d^d \theta^a\, \,\omega
<{\cal {\bf B}}|\theta>\, <\theta|{\cal {\bf C}}> \nonumber\\
&=& \sum^{d}_{k=0} \int %d^{d-1}x
\, b^{*}_{b_{1} \dots b_{k}} c_{b_1 \dots b_{k}}\,.%
\end{eqnarray}%
% from where it follows: $<{\cal {\bf B}}|\theta>=$ $ %<\phi_{og} |
% \sum_{p=0}^{d} (-i)^p\, a^{*}_{a_1\dots a_p} p^{\theta a_p}\,\eta^{a_p a_p}
% \cdots  p^{\theta a_1}\, \eta^{a_1 a_1}$. The vacuum state is chosen to be
% $|\phi_{og}>= |1\,>$, as assumed in Eq.~(\ref{grassmannboson}).

To define norms in Clifford space Eq.~(\ref{grassintegral}) can be used as well.

%Present check and alternatives!!

%***************************************THESE should be left out unless
%*************************************** needed later for scallars or
\section{Expressions for scalar fields in term of $\omega_{s't s}$ and
$\tilde{\omega}_{abs}$ }
\label{scalarsandvectorsapp}

As presented  in Sect.~\ref{scalar3+1} the {\it spin-charge-family}
theory offers the explanation for the appearance of the scalar higss, with the
weak charge $\pm\frac{1}{2}$ and the hyper charge $\mp\frac{1}{2}$, and
Yukawa couplings,  as well as for the matter/antimatter asymmetry in the
universe. There are namely spin  connections of two kinds, $\omega_{abs}$
and $\tilde{\omega}_{abs}$, with the space index $s\ge 5$, which manifest
in $d=(3+1)$ as scalars with the weak charge $\pm\frac{1}{2}$ and the
hyper charge $\mp\frac{1}{2}$ and with the space index $s=(7,8)$,
and there are also scalars which manifest with respect to the space index
$t=(9,10,11,12,13,14)$ as colour triplets or antitriplets.
% Popravi below

\subsection{Scalar fields with the space index $s=(7,8)$}
\label{scalar3+1higgsapp}
The scalar fields, responsible  for masses of the family members and
of the heavy bosons~[18, 19] after gaining
nonzero vacuum expectation values and triggering the electroweak break,
are presented in  the second line of Eq.~(\ref{faction}). These scalar fields
are included in the covariant derivatives of momenta as
$-\frac{1}{2}$ $  S^{s' s"} \omega_{s' s" s} $
$- \frac{1}{2}  \tilde{S}^{a b}  \tilde{\omega}_{a b s}$,
$s \in (7,8)$, $(a,b) $,
$\in (0,\dots,3), (5,\dots,8)$.

One can express the scalar fields carrying the quantum numbers of the
subgroups of the family groups, expressed in terms of
$\tilde{\omega}_{a b s}$ (they contribute to mass matrices of
quarks and leptons and to masses of the heavy bosons), if taking into
account Eqs.~(%
%\ref{so42}, \ref{so64},
\ref{so1+3}, \ref{so42}, \ref{YQY'Q'andtilde}),
%contributing to the electroweak break,
\begin{eqnarray}
\label{Atildeomegas}
\sum_{a,b} -\frac{1}{2}  \tilde{S}^{a b}\,\tilde{\omega}_{a b   s} &=&
- (\vec{\tilde{\tau}}^{\tilde{1}}\, \vec{\tilde{A}}^{\tilde{1}}_{s} +
\vec{\tilde{N}}_{\tilde{L}}\, \vec{\tilde{A}}^{\tilde{N}_{\tilde{L}}}_{s} %\nonumber\\
+ \vec{\tilde{\tau}}^{\tilde{2}}\, \vec{\tilde{A}}^{\tilde{2}}_{s} +
\vec{\tilde{N}}_{\tilde{R}}\, \vec{\tilde{A}}^{\tilde{N}_{\tilde{R}}}_{s})\,,
\nonumber\\
\vec{\tilde{A}}^{\tilde{1}}_{s} &=& (\tilde{\omega}_{5 8 s}-
\tilde{\omega}_{6 7 s},\, \tilde{\omega}_{5 7 s}+
\tilde{\omega}_{6 8 s}, \,\tilde{\omega}_{5 6 s}-
\tilde{\omega}_{7 8 s})\,, \nonumber\\
\vec{\tilde{A}}^{\tilde{N}_{\tilde{L}}}_{s} &=&
(\tilde{\omega}_{2 3 s}+i\,  \tilde{\omega}_{0 1 s}, \,    \tilde{\omega}_{3 1 s}+
i\,  \tilde{\omega}_{0 2 s},  \,   \tilde{\omega}_{1 2 s}+i\,  \tilde{\omega}_{0 3 s})\,, \nonumber\\
\vec{\tilde{A}}^{\tilde{2}}_{s} &=& (\tilde{\omega}_{5 8 s}+
\tilde{\omega}_{6 7 s}, \,\tilde{\omega}_{5 7 s}-
\tilde{\omega}_{6 8 s},\, \tilde{\omega}_{5 6 s}+
\tilde{\omega}_{7 8 s})\,, \nonumber\\
\vec{\tilde{A}}^{\tilde{N}_{\tilde{R}}}_{s} &=& (\tilde{\omega}_{2 3 s}-
i\,  \tilde{\omega}_{0 1 s}, \,    \tilde{\omega}_{3 1 s}-i\,
\tilde{\omega}_{0 2 s}, \,    \tilde{\omega}_{1 2  s}-
i\,  \tilde{\omega}_{0 3 s})\,,\nonumber\\
& &(s\in (7,8))\,.
\end{eqnarray}
Scalars, expressed in terms of $\omega_{abc}$ (contributing as well to
the mass matrices of quarks and leptons and to masses of the heavy bosons) follow, if using
Eqs.~(\ref{so42}, \ref{so64},
%\ref{so1+3tilde}, \ref{so42tilde}) and Eq.~(
\ref{YQY'Q'andtilde})
\begin{eqnarray}
\label{Aomegas}
\sum_{s', s''} \,-\frac{1}{2} S^{s's"}\, \omega_{s's"s}&=&
- (g^{23}\, \tau^{23}\, A^{23}_{s} +
g^{13}\, \tau^{13}\, A^{13}_{s} +
g^{4 }\, \tau^{4} \, A^{4}_{s})\,,
\nonumber\\
g^{13}\,\tau^{13}\, A^{13}_{s} + g^{23}\,\tau^{23}\, A^{23}_{s} + g^{4}\,\tau^{4}\,
A^{4}_{s} &=&
g^{Q}\, Q A^{Q}_{s} + g^{Q'}\, Q' A^{Q'}_{s} + g^{Y'}\,Y'\, A^{Y'}_{s}\,, \nonumber\\
A^{4}_{s} &=& -(\omega_{9\,10\,s} + \omega_{11\,12\,s} + \omega_{13\,14\,s})\,,
\nonumber\\
A^{13}_{s}&=&(\omega_{56 s}- \omega_{78 s})\,, \quad A^{23}_{s}=(\omega_{56 s}+
\omega_{78 s})\,,\nonumber\\
A^{Q}_{s} &=& \sin \vartheta_{1} \,A^{13}_{s} + \cos \vartheta_{1} \,A^{Y}_{s}\,,\quad
A^{Q'}_{s}  = \cos \vartheta_{1} \,A^{13}_{s} - \sin \vartheta_{1} \,A^{Y}_{s}\,,\nonumber\\
A^{Y'}_{s}&=&\cos \vartheta_{2} \,A^{23}_{s} - \sin \vartheta_{2} \,A^{4}_{s}\,,\nonumber\\
& &(s\in (7,8))\,.
%A^{Y}_{s}  =
%g^{56}\,\omega_{56 s} +A^{4}_{s}\,.
\end{eqnarray}
Scalar fields from Eq.~(\ref{Atildeomegas}) interact with quarks and leptons and
antiquarks and antileptons through the family quantum numbers, while those
from Eq.~(\ref{Aomegas})  interact through the family members quantum
numbers. In Eq.~(\ref{Aomegas}) the coupling constants are explicitly written
in order to see the analogy with the gauge fields of the {\it standard model}.

%******Inserted from main

Let be shown that the scalar fields
are {\it triplets} as the gauge fields of the  family quantum numbers
($\vec{\tilde{N}}_{R}, \,$ $\vec{\tilde{N}}_{L},\,$ $ \vec{\tilde{\tau}}^{2},\,$
Eqs.~(\ref{so1+3}, \ref{so42}, \ref{bosonspin0})) or singlets as the gauge fields of
$Q=\tau^{13}+Y, \,Q'= -\tan^{2}\vartheta_{1} Y$ $ + \tau^{13} $ and
$Y' = -\tan^2 \vartheta_{2} \tau^{4} + \tau^{23}$, %~%
for $\tilde{A}^{N_{L}i}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$
and for $A^{Q}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$, taking into account
Eq.~(\ref{so1+3}) (where we replace $S^{ab}$ by ${\cal S}^{ab}$) and
Eq.~(\ref{bosonspin0}), while recognizing that
$\tilde{A}^{N_{L}\spm}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}= $
$\tilde{A}^{N_{L}1}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}\, \smp\, i $
$\tilde{A}^{N_{L}2}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$, and that %further
\begin{eqnarray}
% \label{checktildeNL3Q}
%&=& \spm \tilde{A}^{\tilde{N}_{L}\spm}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}\,,\quad
\tilde{A}^{\tilde{N}_{L}\spm}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}} &=&
\{(\tilde{\omega}_{2 3 \scriptscriptstyle{\stackrel{78}{(\pm)}}} + i \,
\tilde{\omega}_{01 \scriptscriptstyle{\stackrel{78}{(\pm)}}} ) \smp\,i \,
(\tilde{\omega}_{31 \scriptscriptstyle{\stackrel{78}{(\pm)}}}  + i\,
\tilde{\omega}_{02 \scriptscriptstyle{\stackrel{78}{(\pm)}}})\}\,,\nonumber\\
\tilde{A}^{\tilde{N}_{L}3}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}} &=&
(\tilde{\omega}_{12 \scriptscriptstyle{\stackrel{78}{(\pm)}}} +i \,
\tilde{\omega}_{03 \scriptscriptstyle{\stackrel{78}{(\pm)}}})\,, \nonumber\\
% %Q \,A^{Q}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}} &=&0\,,
A^{Q}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}} &=&
\omega_{56 \scriptscriptstyle{\stackrel{78}{(\pm)}}} -
(\omega_{9\,10 \scriptscriptstyle{\stackrel{78}{(\pm)}}} +
\omega_{11\,12 \scriptscriptstyle{\stackrel{78}{(\pm)}}} +
\omega_{13\,14 \scriptscriptstyle{\stackrel{78}{(\pm)}}})\,, \nonumber
\end{eqnarray}
the scalar fields are chosen (with respect to the space index $s=(7,8)$) to be
eigenstates of the weak charge, Eq.~(\ref{eigentau1tau2}).

One finds
\begin{eqnarray}
\label{checktildeNL3Q1}
\tilde{N}_{L}^{3}\,
\tilde{A}^{\tilde{N}_{L}\spm}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}
&=& \spm \tilde{A}^{\tilde{N}_{L}\spm}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}\,,
\quad \tilde{N}_{L}^{3}\,
\tilde{A}^{\tilde{N}_{L}3}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}=0
\,,\nonumber\\
Q \,A^{Q}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}} &=&0\,,
\end{eqnarray}
%One applies $\tilde{\tau}^{23}$,  $\tilde{\tau}^{13}$, $\tilde{N}_{R}^{3}$
%and $\tilde{N}_{L}^{3}$ on their eigenstates.
taking into account $ Q={\cal S}^{56} + {\cal \tau}^{4}= {\cal S}^{56} -
\frac{1}{3}({\cal S}^{9\,10}+
{\cal S}^{11\,12} + {\cal S}^{13\,14})$, and with
${\cal \tau}^{4}$ defined in Eq.~(\ref{so64}), if replacing
$S^{ab}$ by ${\cal S}^{ab}$ from Eq.~(\ref{bosonspin0}).
Similarly one finds properties with respect to the $Ai$ quantum
numbers for all the scalar fields $A^{Ai}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$.
% \end{small}
\subsection{Scalar fields with the space index $s=(9,\dots,14)$}
\label{scalar3+1matterantimatterapp}

%******Inserted from main
%*Correct 05.02.2021 at 18:18\\
%%%%    29.11.2020 at 18:03.  Matter/antimatter

Expressions for the vector gauge fields in terms of the spin connection fields
and the vielbeins, which correspond to the colour charge ($\vec{A}^{3}_{m}$),
the $SU(2)_{II}$ charge ($\vec{A}^{2}_{m}$), the weak $SU(2)_{I}$ charge
($\vec{A}^{1}_{m}$) and
the $U(1)$ charge originating in $SO(6)$ ($\vec{A}^{4}_{m}$), can be found
by taking into account Eqs.~(\ref{so42}, \ref{so64}). Equivalently one finds
the vector gauge fields in the "tilde" sector, or one just uses the expressions
from Eqs.~(\ref{Aomegas}, \ref{Atildeomegas}), if replacing the
scalar index $s$ with the vector index $m$.

The expression for  $\sum_{t a b}\gamma^{t}\,$
$\frac{1}{2}\, \tilde{S}^{ab} \,\tilde{\omega}_{ab t}$, used in
($\tilde{S}^{ab}$ are the infinitesimal generators of either $\widetilde{SO}(3,1)$ or
$\widetilde{SO}(4)$, while $\tilde{\omega}_{ab t}$ belong to the corresponding
gauge fields  with $t =(9,\dots,14)$), and
obtained by using Eqs.~(\ref{so1+3} - \ref{plusminus}), are
\begin{eqnarray}
&&\sum_{a b t}\,\gamma^{t} \frac{1}{2}\, \tilde{S}^{ab} \,\tilde{\omega}_{ab t} =
\sum_{+ - t t' a b}\,\stackrel{t t'}{(\cpm)} \, \frac{1}{2}\, \tilde{S}^{a b} \,
\tilde{\omega}_{\scriptscriptstyle{ab \,\stackrel{t t'}{(\cpm)}}}= \nonumber\\
&&\sum_{+ - t t'}\stackrel{t t'}{(\cpm)} \,
\{\,\tilde{\tau}^{2+}\,\tilde{A}^{2+}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} +
\tilde{\tau}^{2-}\,\tilde{A}^{2-}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} +
\tilde{\tau}^{23}\,\tilde{A}^{23}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} + \nonumber\\
& & \tilde{\tau}^{1+}\,\tilde{A}^{1+}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} +
\tilde{\tau}^{1-}\,\tilde{A}^{1-}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} +
\tilde{\tau}^{13}\,\tilde{A}^{13}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} + \nonumber\\
& & \tilde{N}^{+}_{R}\,\tilde{A}^{N_{R}+}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} +
\tilde{N}^{-}_{R}\,\tilde{A}^{N_{R}-}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} +
\tilde{N}^{3}_{R}\,\tilde{A}^{N_{R}3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} + \nonumber\\
& &
\tilde{N}^{+}_{L}\,\tilde{A}^{N_{L}+}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} +
\tilde{N}^{-}_{L}\,\tilde{A}^{N_{L}-}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} +
\tilde{N}^{3}_{L}\,\tilde{A}^{N_{L}3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}\,\}
\,,\nonumber\\
\tilde{A}^{N_{R}\spm}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} &=&
(\tilde{\omega}_{\scriptscriptstyle{23 \stackrel{t t'}{(\cpm)}}}- i\,
\tilde{\omega}_{\scriptscriptstyle{01 \stackrel{t t'}{(\cpm)}}})\smp \, i
(\tilde{\omega}_{\scriptscriptstyle{31 \stackrel{t t'}{(\cpm)}}}- i\,
\tilde{\omega}_{\scriptscriptstyle{02 \stackrel{t t'}{(\cpm)}}})\,,\quad
\tilde{A}^{N_{R}3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}=
(\tilde{\omega}_{\scriptscriptstyle{12 \,\stackrel{t t'}{(\cpm)}}}- i\,
\tilde{\omega}_{\scriptscriptstyle{03 \stackrel{t t'}{(\cpm)}}})\,, \nonumber\\
\tilde{A}^{N_{L}\spm}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}} &=&
(\tilde{\omega}_{\scriptscriptstyle{23 \stackrel{t t'}{(\cpm)}}}+ i\,
\tilde{\omega}_{\scriptscriptstyle{01 \stackrel{t t'}{(\cpm)}}})\smp \, i
(\tilde{\omega}_{\scriptscriptstyle{31 \stackrel{t t'}{(\cpm)}}}+ i\,
\tilde{\omega}_{\scriptscriptstyle{02 \stackrel{t t'}{(\cpm)}}})\,,\quad
\tilde{A}^{N_{R}3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}=
(\tilde{\omega}_{\scriptscriptstyle{12 \stackrel{t t'}{(\cpm)}}}+ i\,
\tilde{\omega}_{\scriptscriptstyle{03 \stackrel{t t'}{(\cpm)}}})\,.
\label{factionMaMpart20}
\end{eqnarray}
% The expressions for $\tilde{A}^{2\spm}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$,
%  $\tilde{A}^{23}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$,
% $\tilde{A}^{1\spm}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$ and
%  $\tilde{A}^{1 3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$ can easily be obtained
%  from Eq.(\ref{factionMaMpart11}) by replacing in expressions  for
%  $A^{2\spm}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$,
%  $A^{23}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$,
% $A^{1\spm}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$
% and $A^{1 3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$, respectively,
%%(Eq.(\ref{factionMaMpart11})), if replacing
% $\omega_{\scriptscriptstyle{s' s" \,\stackrel{t t'}{(\cpm)}}}$ by
% $\tilde{\omega}_{\scriptscriptstyle{s' s" \,\stackrel{t t'}{(\cpm)}}}$.
%%In an equivalent
%%way also the expressions for $\tilde{A}^{N_{R}\spm}_{\scriptscriptstyle{\stackrel{t t'}% %{(\cpm)}}}$,
% %$\tilde{A}^{N_{R}3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$ and
% %$\tilde{A}^{N_{L}\spm}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$,
% %$\tilde{A}^{N_{L}3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$ can be found.

The term $\sum_{t t' t''}\gamma^t$ $\frac{1}{2}\,S^{t' t"}\,\omega_{t' t" t}$ in Eq.~(\ref{faction})
% and $\gamma^m$ $\frac{1}{2}\,S^{t' t"}\,\omega_{t' t" m}$.
%$\gamma^t$ $\frac{1}{2}\, S^{t' t"}\,\omega_{t' t" t}$
can be rewritten  with respect to the generators $S^{t' t"}$
% **, similarly as the terms
% $\gamma^t$ $\frac{1}{2}\, S^{s' s"}\,\omega_{s' s" t}$ and $\gamma^t$
%$\frac{1}{2}\, \tilde{S}^{s' s"} $
and the corresponding gauge fields $\omega_{s' s" t}$
as one colour octet scalar field and one $U(1)_{II}$ singlet scalar field (Eq.~\ref{so64})
\begin{eqnarray}
\sum_{t t'' t'''}\gamma^{t}\,\frac{1}{2}\, S^{t" t'"} \,\omega_{t" t'" t} &=&
\sum_{+,-}\,\sum_{(t\,t')}  \, \stackrel{t t'}{(\cpm)}\,
\{\,\vec{\tau}^{3}\cdot \vec{A}^{3}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}\, + \tau^{4}\cdot
A^{4}_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}\,\}\,, \nonumber\\
(t\,t') &\in& ((9\,10), 11\,12),13\,14))\,.
\label{factionMaMpart21}
\end{eqnarray}

%  Considering all the above equations~(\ref{factionMaMpart10} -
%\ref{factionMaMpart11}, \ref{factionMaMpart20},
%  \ref{factionMaMpart21}),  %Eq.~(\ref{faction}) can be rewritten,
% and leaving out $p_{\scriptscriptstyle{\stackrel{t t'}{(\cpm)}}}$
% since in the low energy limit the momentum does not play any role, the
% ${\mathcal L}_{f" }$  follows

\section{Handedness in Grassmann and Clifford space}
\label{handednessGrassCliff}

The handedness $\Gamma^{(d)}$ is one of the invariants of the group $SO(d)$,
with the infinitesimal generators of the Lorentz group $S^{ab}$,
defined as
\begin{eqnarray}
\label{handedness}
\Gamma^{(d)}&=&\alpha\, \varepsilon_{a_1 a_2\dots a_{d-1} a_d}\, S^{a_1 a_2}
\cdot S^{a_3 a_4} \cdots S^{a_{d-1} a_d}\,,
\end{eqnarray}
with $\alpha$, which is chosen so that $\Gamma^{(d)}=\pm 1$.

In the Grassmann case  $S^{ab}$  is defined in Eq.~(\ref{thetasab}), while in the Clifford case
Eq.~(\ref{handedness}) simplifies, if we take into account that $S^{ab}|_{a\ne b}=
\frac{i}{2}\gamma^a \gamma^b$  and $\tilde{S}^{ab}|_{a\ne b}=
\frac{i}{2}\tilde{\gamma}^a \tilde{\gamma}^b$, as follows
\begin{eqnarray}
\Gamma^{(d)} :&=&(i)^{d/2}\; \;\;\;\;\;\prod_a \quad (\sqrt{\eta^{aa}} \gamma^a),
\quad {\rm if } \quad d = 2n\,.
\nonumber\\
%\Gamma^{(d)} :&=& (i)^{(d-1)/2}\; \prod_a \quad (\sqrt{\eta^{aa}} \gamma^a),
%\quad {\rm if } \quad d = 2n +1\,.
\label{hand}
\end{eqnarray}

\section{Discrete symmetries of vector and scalar gauge fields in $d=(3+1)$  from
$d=((d-1) +1)$~[136]}
\label{DSVS}

% *TU 16.04. at 15: 45 This is taken from ~[136]   AND MUST BE SHORTENED,
% AND PRESENTED MORE PEDAGOGICALY , 10.03.2021, corrected 17.04.2021

%\subsection{Interacting spinors}
We follow here Ref.~[136].
In Sect.~\ref{CPT} our definition~[136] of the discrete symmetry operators
$\mathbb{C}_{{\cal N}}$, ${\cal P}^{(d-1)}_{{\cal N}} $ and $ \cal{T}_{{\cal N}}$,
Eq.~(\ref{CPTNlowE}), is presented and their application on the second quantized states of
fermions and antifermions, living in $d=((d-1)+1)$ spaces and manifesting in $d=(3+1)$
as the observed fermions and antifermions, is discussed.
%Independent of how their internal space is described, with the odd Clifford algebra
%of $\gamma^a$'s or not.

In this App.~\ref{DSVS} we apply  the discrete symmetry operators
$\mathbb{C}_{{\cal N}}$, ${\cal P}^{(d-1)}_{{\cal N}} $ and $ \cal{T}_{{\cal N}}$,
Eq.~(\ref{CPTNlowE}), on vielbeins and spin connection fields, living in
$d=((d-1)+1)$ spaces and manifesting  $d=(3+1)$ as the vector and scalar gauge
fields (as it is in the Kaluza-Klein theories)~[17].
We again follow Ref.~[136].

Let us treat the simple starting Lagrange density for a spinor in $d=((d-1) +1)$
dimensional space, which carries, like in the Kaluza-Klein theories, the spins and no
charges, Eqs.~(\ref{wholeaction}, \ref{faction}), Sect.~\ref{fermionandgravitySCFT}
% Let it in addition carry the families quantum numbers~\cite{norma,Portoroz03,%
% pikanorma,NPLB,NJMP}.
\begin{eqnarray}
\label{lagrangespinsfamilies0}
{\cal L_f} &=&\frac{1}{2} \,E \, \Psi^{\dagger} \, \gamma^0 \, \gamma^a\, p_{0a}\,\Psi + \, h.c.\,,\nonumber\\
p_{0a}   &=& f^{\alpha}_{a} p_{\alpha} + \frac{1}{2E} \{p_{\alpha},f^{\alpha}_{a} E \}_{-}-
\frac{1}{2}\, S^{cd}\, f^{\alpha}_{a}\,\omega_{cd \alpha}\,.
%-\frac{1}{2}\, \tilde{S}^{cd}\, f^{\alpha}_{a}\,\tilde{\omega}_{cd \alpha}\,.
\end{eqnarray}
$f^{\alpha}_{a}$ are vielbein and $\omega_{cd \alpha}$ spin connection fields, the gauge fields of
$p^{a}$ and $S^{ab}$, respectively.
%In this paper we do not discuss
The families quantum numbers, determined by $\tilde{S}^{ab}$,
commute, Eq.~(\ref{gammatildeantiher}), with in Sect.~\ref{CPT} defined
discrete symmetries operators~\footnote{
The discrete symmetry operators for the "Grassmann fermions" are presented in Refs.~%
[23, 24, 26]. }.
% and  $\tilde{\omega}_{cd \alpha}$ are the gauge fields of $\tilde{S}^{ab}$, with
%\{S^{ab}, \tilde{S}^{ab}\}_{-} &=&0\,.
% This should go into appendix.%%%A

Let the vielbeins and spin connections be responsible for the break of symmetry of
$M^{(d-1)+1}$ into $M^{3+1}\times$ $M^{d-4}$ so that the manifold $M^{d-4}$
is (almost) compactified~[138, 139, 140] and let the spinors (fermions)
manifest in $d=(3+1)$ the ordinary spin and the charges and the families
~\footnote{In Refs.~[138, 139, 140] it is demonstrated on the
toy model how such an almost compactification could occur.}.
Looking for the subgroups (denoted by $A$)  of the  $SO((d-1)+1)$
%and $\tilde{SO}((d-1)+1)$ (denoted by $D$)
group and assuming no gravity in $d=(3+1)$, the Lagrange density of
Eq.~(\ref{lagrangespinsfamilies0}) can be rewritten in a more familiar shape
\begin{eqnarray}
\label{lagrangespinsfamilies1}
{\cal L} &=& \frac{1}{2} \,E \, \psi^{\dagger} \, \gamma^0 \, (\gamma^m \,
p_{0m} + \gamma^s \, p_{0s})\,\,\psi \,+ h.c. \,,\nonumber\\
p_{0m}   &=&  p_{m} - \sum_{A}\,\vec{\tau}^{A}\,\vec{A}^{A}_{m}\,,\nonumber\\
p_{0s}   &=&  f^{\sigma}_{s} \,p_{\sigma} + \frac{1}{2E} \{p_{\sigma},
f^{\sigma}_{s} E \}_{-} - \sum_{B}\, \vec{\tau}^{B}\,\vec{A}^{B}_{s}\,,
%- \sum_{D}\, \vec{\tilde{\tau}}^{D}\,\tilde{\vec{A}}^{D}_{s}\,,
\end{eqnarray}
with $m=(0,1,2,3)$,  $s=(5,6,\dots,d)$ and
$\tau^{Ai}=        \sum_{st}\, c^{Ai}{}_{st}\, S^{st}\,$, %\nonumber\\
%$\tau^{Ci}=        \sum_{st}\, c^{Ci}{}_{st}\, S^{st}\,$, %\nonumber\\
%$\tilde{\tau}^{Di}=\sum_{ab}\, \tilde{d}^{Di}{}_{ab}\, \tilde{S}^{ab}\,$,
$\sum_{A}\,\vec{\tau}^{A}\,\vec{A}^{A}_{m}= \frac{1}{2}\, \sum_{s t}\,S^{st}\,
\omega_{st m}\,$,
$\sum_{A}\,\vec{\tau}^{A}\,\vec{A}^{A}_{s}= \frac{1}{2}\,\sum_{st}\, S^{st}\,
f^{\sigma}_{s} \omega_{st \sigma}\,$.
%$\sum_{D}\,\vec{\tilde{\tau}}^{D}\,\vec{A}^{D}_{s}= \frac{1}{2}\, \tilde{S}^{ab}\,
% f^{\sigma}_{s}
%\tilde{\omega}_{ab \sigma}$.

Taking into account Eqs.~(\ref{so42}, \ref{so64}) (and recognizing that all $\tau^{Ai}
=\sum_{A,i} c^{Ai}{}_{st} \,S^{st}$ include only $S^{s t}= \frac{i}{2} \gamma^s
\gamma^t $ with either $\gamma^s$ or $\gamma^t$ imaginary and the remaining
one is real, Eq.~(\ref{complexgamatilde})) one easily finds  that
\begin{eqnarray}
\label{CNtransempt}
\mathbb{C}_{\cal N}\,\,  \tau^{Ai}\,\, \mathbb{C}_{\cal N}^{-1} &=& - \tau^{Ai}\,,
\nonumber\\
\mathbb{C}_{\cal N}{\cal P}^{(d-1)}_{\cal N}\,\,  \tau^{Ai}\,\, (\mathbb{C}_{\cal N}
{\cal P}^{(d-1)}_{\cal N})^{-1} &=& - \tau^{Ai}\,, \nonumber\\
\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N}\,\, \tau^{Bi}
%\,A^{Bi}_{m} (x)
\,(\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N})^{-1}
&=& -\tau^{Bi}\,.
\end{eqnarray}
Taking into account  how do the vector gauge  fields transform according to prescription
in Sect.~\ref{CPT}, Eq.~(\ref{CPTNlowE}), one further finds
\begin{eqnarray}
\label{CNtransempt1}
\mathbb{C}_{\cal N}\, \, A^{Ai}_{m}(x^0,\vec{x}_3)\,\, \mathbb{C}_{\cal N}^{-1}
&=& -A^{Ai}_{m}(x^0,\vec{x}_3)\,, \nonumber\\
\mathbb{C}_{\cal N}{\cal P}^{(d-1)}_{\cal N}\,\,  A^{Ai}_{m}(x^0,\vec{x}_3) \,\,
(\mathbb{C}_{\cal N}{\cal P}^{(d-1)}_{\cal N})^{-1} &=&
- A^{Ai\,m}(x^0,-\vec{x}_3)\,,\nonumber\\
\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N}\,\, \tau^{Ai}\,
A^{Ai}_{m} (x)\,\,
(\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N})^{-1}
&=& (-\tau^{Ai})\,(-A^{Ai*}_{m}(-x))\,,
\end{eqnarray}
for $\tau^{Ai}$  from the Cartan subalgebra for each $A$. It  is always true that
transforms either to $(-\tau^{Ai})\,(-A^{Ai}_{m})$  or to $\tau^{Ai}\,A^{Ai}_{m}$,
for each $Ai$,
all in agreement with the standard knowledge for the gauge vector fields and charges
in $d=(3+1)$~[145].

One  can  check that
\begin{eqnarray}
\label{fieldsproperties}
\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N}\, \gamma^a
(\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N})^{-1}&=&
\gamma^a \,,\nonumber\\
%$(-)\,$ for $a=m =(0,\dots,3)\,$, $(+)$ for $a=s =(5,6,\dots,d)\,$; %\nonumber\\
\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N}  S^{ab}\,
(\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N})^{-1}&=&- S^{ab}\,,
\nonumber\\
%$(-)$ for $(a,b)=(m,n)$ or $(a,b)=(s,t)\,$, $(+)$ for $(a,b)=(m,s)\,$;
%$\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N} \,\tilde{S}^{ab}\,$
%$(\mathbb{C}_{\cal N} {\cal P}_{\cal N} {\cal T}_{\cal N})^{-1}= (\mp)\,
%$(-)$ for $(\tilde{\gamma}^a,\tilde{\gamma}^b) \in(\Im \tilde{\gamma^a},
% \Im \tilde{\gamma^b})\,$ or
%$(\tilde{\gamma}^a,\tilde{\gamma}^b)$ $\in(\Re \tilde{\gamma^a},
%\Re \tilde{\gamma^b})\,$, %\nonumber\\
%$(+)$  otherwise; %\nonumber\\
\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N} \,
(\mathbb{C}_{\cal N} {\cal P}_{\cal N} {\cal T}_{\cal N})^{-1} &=&
\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N} \omega_{abc}(x)\,
(\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N})^{-1}
&=& - \omega^{*}_{abc}(-x) \,,\nonumber\\
\mathbb{C}_{\cal N} {\cal P}_{\cal N} {\cal T}_{\cal N}\, \tilde{\omega}_{abc} (x)\,
(\mathbb{C}_{\cal N} {\cal P}_{\cal N} {\cal T}_{\cal N})^{-1}&=&
- \tilde{\omega}^{*}_{abc} (-x) \,.
\end{eqnarray}
%From where it follows that operators representing spins and charges and the corresponding gauge vector and scalar
%fields (with respect to $d=(3+1)$) behave with respect to $\mathbb{C}_{\cal N} $
%${\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N} $ in a standard way~\cite{ItZu}, leading to
%let it go to the appendix
%$\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N}\,\tau^{Bi}\,A^{Bi}_{m} (x)
%(\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N})^{-1} $
%$= (-\tau^{Bi})\,(-A^{Bi*}_{m}(-x))\,$,

From Eqs.~(\ref{CNtransempt}, \ref{CNtransempt1}, \ref{fieldsproperties}) it follows
$$\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N}\,\tau^{Ai}\,
A^{Ai}_{s} (x) (\mathbb{C}_{\cal N} {\cal P}^{(d-1)}_{\cal N} {\cal T}_{\cal N})^{-1}
= (-\tau^{Ai})\,(-A^{Ai*}_{s})(-x))\,,$$
concerning in $d=(3+1)$ the gauge scalar fields,  which determine in
the {\it spin-charge-family} theory massless and massive solutions for fermions
(quarks and leptons and antiquarks and antileptons) and, when some
of them gaining  constant values --- the one with the space
index $s=(7,8)$ --- contribute not only  to masses of
spinors but also to those gauge fields, to which they couple.

The Hermiticity requirement for the Lagrange density (Eq.~(\ref{lagrangespinsfamilies1})),
${\cal L}^{\dagger}= {\cal L}$, leads to
\begin{eqnarray}
\label{hcreq}
\omega^{*}_{abc}(x)&=& (\mp)\,\omega_{abc}(x)\,;(-)\, {\rm if}\, a=c \,{\rm or}\, b=c\,,
(+)\, {\rm otherwise}\,,
%\tilde{\omega}^{*}_{abc}(x)&=& (\mp)\,\tilde{\omega}_{abc}(x)\,,(-)\,, {\rm if}\, a=0 \,{\rm or}\, b=0\,,
%(+)\,, {\rm otherwise}\,.
\end{eqnarray}
which is to be taken into account together with the $\mathbb{C}_{\cal N} {\cal P}_{\cal N} {\cal T}_{\cal N}$

For additional arguments about the generality of our proposal for the discrete symmetry
operators $\mathbb{C}_{\cal N}$, ${\cal P}^{(d-1)}_{\cal N}$ and ${\cal T}_{\cal N}$
the reader is invited to follow Ref.~[136].

\section{Symmetries of mass matrices on  tree level and beyond
manifesting  $SU(2)\times SU(2) \times U(1)$ symmetry in  {\it
spin-charge-family} theory~[104, 105]}
\label{M0SCFT}

Let states of massless quarks and leptons for two times four families ---
presented in Table~\ref{Table III.} with creation operators, creating the
internal part of the right handed  $u$-quark and right handed
$\nu$-leptons if applying on the vacuum state $|\psi_{oc}>$ ---
be denoted by $\psi^{\alpha}_{f}$, $\alpha=(u,d,\nu, e)_{L,R}$ and
$f$ is the family quantum number.
Table~\ref{Table so13+1.} represents all the members of the first family
of Table~\ref{Table III.}. All the members of any of two times four families
follows by the application of $S^{ab}$.

The scalar gauge fields ($\vec{\tilde{\tau}}^{1}\,\cdot
\vec{\tilde{A}}^{\tilde{1}}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$,
$ \vec{\tilde{N}}_{L}\, \cdot
\vec{\tilde{A}}^{\tilde{N}_{\tilde{L}}}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$)
with the space index $s=(7,8)$ transform the right handed members into the
left handed ones, causing as well transformations among families
of the first group. %, multiplaying the second group of four families with zero.

The scalar gauge fields ($\vec{\tilde{\tau}}^{2}\,\cdot
\vec{\tilde{A}}^{\tilde{2}}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$,
$ \vec{\tilde{N}}_{R}\,\cdot
\vec{\tilde{A}}^{\tilde{N}_{\tilde{R}}}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$)
behave equivalently on the second group of the four families.

The scalar gauge fields ($Q \cdot A^{Q}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}},
Y \cdot A^{Q]Y}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}},
\tau^4 \cdot A^{4}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$) "see" only the
family members quantum numbers, independently of the family quantum number.
They are diagonal and apply on both groups of four families equivalently.

Although the two groups of scalar fields might have different properties, different
different masses and different coupling constants, they bring to the two groups
of four families the equivalent
$SU(2)\times SU(2) \times U(1)$ symmetry~%
\footnote{The infinitesimal generators $\tilde{N}^{i}_{L}\,,\,i=(1,2,3)$ and
$\tilde{N}^{i}_{R}\,,\,i=(1,2,3)$  determine the algebra
of the two invariant subgroups of the $\widetilde{SO}(1,3)$ group, while
$\tilde{\tau}^{1i}\,,\,i=(1,2,3)$ and $\tilde{\tau}^{2i}\,,\,i=(1,2,3)$ determine
the two invariant subgroups of the $\widetilde{SO}(4)$ group. The first four
families of Table~\ref{Table III.}, carrying the quantum numbers of
$\tilde{N}^{i}_{L}$ and $ \tilde{\tau}^{1i}$, are influenced by the
corresponding scalar gauge fields, the second  four families are influenced by
the scalar fields of $\tilde{N}^{i}_{R}$ and $ \tilde{\tau}^{2i}$.}

Let us therefore treat only one group of four families, the first one, interacting with
\vec{\tilde{A}}^{\tilde{1}}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$,
$ \vec{\tilde{N}}_{L}\, \cdot
\vec{\tilde{A}}^{\tilde{N}_{\tilde{L}}}_{\scriptscriptstyle{\stackrel{78}{(\pm)}}}$).
The operators
\begin{eqnarray}
\label{taunl}
\tilde{N}^{i}_{L}\,,\,i=(1,2,3)\,, && {\tau}^{1i}\,,\,i=(1,2,3)\,,
\nonumber\\
\{\tilde{N}^{i}_{L}, \tilde{N}^{j}_{L}\}_{-}= i\,\varepsilon^{ijk} \tilde{N}^{k}_{L}
\,,\;\;
\{\tilde{\tau}^{1i},\tilde{\tau}^{1j}\}_{-}&=& i\,\varepsilon^{ijk} \tilde{\tau}^{1k}
\,,\{\tilde{N}^{i}_{L}, \tilde{\tau}^{1j}\}_{-}=0\,,
\end{eqnarray}
where $\varepsilon^{ijk}$ is the totally antisymmetric tensor, transform the basic
vectors $\psi^{\alpha}_{i}$, into one another  as follows
\begin{eqnarray}
\label{taunlonpsi}
&&\tilde{N}^{3}_{L}\, (\psi^{\alpha}_1, \psi^{\alpha}_2,\psi^{\alpha}_3,
\psi^{\alpha}_4)=
\frac{1}{2}\, (-\psi^{\alpha}_1, \psi^{\alpha}_2,-\psi^{\alpha}_3,
\psi^{\alpha}_4)\,,\nonumber\\
&&\tilde{N}^{+}_{L}\, (\psi^{\alpha}_1, \psi^{\alpha}_2,\psi^{\alpha}_3,
\psi^{\alpha}_4)=
(\psi^{\alpha}_2, \;\;0,\psi^{\alpha}_4,\;\;0)\,,\nonumber\\
&&\tilde{N}^{-}_{L}\, (\psi^{\alpha}_1, \psi^{\alpha}_2,\psi^{\alpha}_3,
\psi^{\alpha}_4)=
(0\;\;, \psi^{\alpha}_1,\;\;0,\psi^{\alpha}_3)\,,\nonumber\\
&&\tilde{\tau}^{13}\, (\psi^{\alpha}_1, \psi^{\alpha}_2,\psi^{\alpha}_3,
\psi^{\alpha}_4)=
\frac{1}{2}\, (-\psi^{\alpha}_1, -\psi^{\alpha}_2,\psi^{\alpha}_3,
\psi^{\alpha}_4)\,, \nonumber\\
&&\tilde{\tau}^{1+}\, (\psi^{\alpha}_1, \psi^{\alpha}_2,\psi^{\alpha}_3,
\psi^{\alpha}_4)=
(\psi^{\alpha}_3, \psi^{\alpha}_4,\;\;0,\;\;0)\,,\nonumber\\
&&\tilde{\tau}^{1-}\, (\psi^{\alpha}_1, \psi^{\alpha}_2,\psi^{\alpha}_3,
\psi^{\alpha}_4)=
(\;\;0, \;\;0,\psi^{\alpha}_1,\psi^{\alpha}_2) \,.
\end{eqnarray}
The three $U(1)$ operators ($Q,Y$ and $\tau^4$) (or any three superposition of
them, like $Q, Q', Y'$) commute with the family operators
and $\vec{\tilde{\tau}}^{1}$, distinguishing only among family
members $\alpha$.

%*04.03.2020  at 14:20\\

\begin{eqnarray}
\label{QQ'Y'nltau}
&&\{ \tilde{N}^{i}_{L}\,, (Q,Q',Y') \}_{-} =(0,0,0)\,,\nonumber\\
&&\{ \tilde{\tau}^{1i}\,, (Q,Q',Y') \}_{-} =(0,0,0)\,,\nonumber\\
&&(Q\,, Y\,, \tau^4)\, (\psi^{\alpha}_1, \psi^{\alpha}_2,\psi^{\alpha}_3,
\psi^{\alpha}_4)=
(Q^{\alpha}\,, Y^{\alpha}\,, \tau^{4 \alpha} )\,
(\psi^{\alpha}_1, \psi^{\alpha}_2,\psi^{\alpha}_3,\psi^{\alpha}_4) \,,
\end{eqnarray}
giving the same eigenvalues for all the families.

The nonzero vacuum expectation values of the  scalar gauge fields of
$\tilde{N}^{i}_{L}$  ($\tilde{A}^{\tilde{N}_{L}i}_{s}=$ $
\sum_{\tilde{m}\tilde{n}}\tilde{c}^{\tilde{N}_{L}i}{}_{\tilde{m}\tilde{n}}\,
\tilde{\omega}^{\tilde{m}\tilde{n}}{}_{s}$, $(\tilde{m}, \,\tilde{n}) =(0,1,2,3)$),
of  $\tilde{\tau}^{1i}$  ($\tilde{A}^{\tilde{1}i}_{s}= \sum_{\tilde{t}\tilde{t'}}
\tilde{c}^{\tilde{1}i}{}_{\tilde{t} \tilde{t'}}\, \tilde{\omega}^{\tilde{t}\tilde{t'}}{}_{s}
$,  $(\tilde{t},\tilde{t'}) =(5,6,7,8)$) and of the three singlet scalar gauge fields of
($Q,Y$ and $\tau^4$), which all are
superposition of $\omega_{t,t',s}$ ($A^{Q}_{s} = \sum_{t\, t'} c^{Q}{}_{t t'}\,
\omega^{t t'}{}_{s}$\,,
$A^{Y}_{s} =  \sum_{t\, t'} c^{Y}{}_{t t'}\,\omega^{t t'}{}_{s}$ and
$A^{4}_{s} =  \sum_{t\, t'} c^{Y'}{}_{t t'}\,
\omega^{t t'}{}_{s}$, $s=(7,8)$, $(t,t') =(5,6,7,8)$), determine on the tree level,
together with the
corresponding coupling  constants,  the
$SU(2) \times SU(2) \times U(1)$ symmetry and the strength of the mass matrix
of each family member $\alpha$, Eq.~(\ref{M0}).
%The nonzero vacuum expectation values of the
%scalar fields $A^{Q}_{s}, A^{Q'}_{s},A^{Y'}_{s},$ the gauge scalar fields of $Q,Q',Y'$,
% which are the superposition of $\omega_{t t's}$, contribute on the tree level to diagonal
% terms distinguishing among
%family members only.
In loop corrections  the scalar fields
$ \tilde{A}^{\tilde{N}_{L}i}_{s}$,  $\tilde{A}^{\tilde{1}i}_{s}$,
$A^{Q}_{s}, A^{Y}_{s},A^{4}_{s}$  contribute to all the matrix elements,
keeping the symmetry unchanged, Eq.~(\ref{M0}). The twice  two zeros on the
tree level obtain in loop corrections the value $b$.

One easily checks that a change of the phases of the left and the right
handed members, there are $(2n-1)$
possibilities, causes  changes in phases of matrix elements in Eq.~(\ref{M0}).

All the scalars are doublets with respect to the weak charge,
contributing to the weak and the hyper charge of the fermions so that they
transform the right handed members into
the left handed onces, Sect.~\ref{scalar3+1}, %~\cite{N2014scalarprop},
what is in the {\it standard model} just

%Let us, as one example,
%change the phases of the right
%handed basis as follows: $(\psi_1, \psi_2,\psi_3,\psi_4)$ into $(\psi_1, \psi_2,\psi_3,-\psi_4)$.
%Then the matrix~Eq.(\ref{M0}) changes  as follows
% \begin{equation}
% \label{M0phase4}
% {\cal M}^{\alpha} = \begin{pmatrix}
% - a_1 - a & e     &   d &{\bf -} b\\
% e     & - a_2 - a &   b & {\bf -}d\\
% d     & b     & a_2 - a & {\bf -}e\\
% {\bf -} b     & {\bf -} d    & {\bf -} e   & {\bf -}(a_1 - a)
% \end{pmatrix}\,.
% \end{equation}
%Let us point out that on the tree level $b_1=b_2=0$. Loop corrections change these two zeros into $b_1$ and
%$b_2$, which might even be equal. \\
%% Paper!!(**Check whether or not $b_1=b_2$.**)

%*Below can be omitted
\subsection{Properties of non Hermitian mass matrices}
\label{nonhermitean}
% O.K.

This pedagogic presentation of well known properties of non Hermitian matrices can be
found in many textbooks. %, for example~\cite{chine}.
We repeat this topic here only to make our discussions  transparent.

Let us take   a  non Hermitian mass matrix  $M^{\alpha}$ as it follows from the
{\it spin-charge-family} theory, $\alpha$ denotes a family member.
% (index ${}_{\pm}$ used in the main text is dropped).

We always can diagonalize  a non Hermitian $M^{\alpha}$ with two unitary matrices,
$S^{\alpha}$ ($S^{\alpha\, \dagger}\,S^{\alpha}=I$)
and $T^{\alpha}$ ($T^{\alpha\, \dagger}\,T^{\alpha}=I$)
\begin{eqnarray}
\label{diagnonher0}
S^{\alpha\, \dagger}\,M^{\alpha}\,T^{\alpha}&=& {\bf M}^{\alpha}_{d}\,=
(m^{\alpha}_{1}\, \dots m^{\alpha}_{i}\, \dots m^{\alpha}_{n}).
\end{eqnarray}
The proof is added below.

Changing phases of the basic states, those of the left handed one and those
of the right handed one, the new unitary matrices  $S'^{\alpha} =
S^{\alpha} \,F_{\alpha S} $ and $T'^{\alpha} = T^{\alpha}\,F_{\alpha T}$
change the phase of the elements of diagonalized mass matrices
${\bf M}^{\alpha}_{d}$
\begin{eqnarray}
\label{diagnonher}
S'^{\alpha\, \dagger}\,M^{\alpha}\,T'^{\alpha}&=& F^{\dagger}_{\alpha S}\,{\bf M}^{\alpha}_{d}\,
F_{\alpha T}=\nonumber\\
& & diag(m^{\alpha}_{1} e^{i(\phi^{\alpha S}_{1}- \phi^{\alpha T}_{1})}\, \dots
m^{\alpha}_{i}\, e^{i(\phi^{\alpha S}_{i}- \phi^{\alpha T}_{i})}\,, \dots m^{\alpha}_{n}\,
e^{i(\phi^{\alpha S}_{n}- \phi^{\alpha T}_{n})})\,,\nonumber\\
F_{\alpha S} &=& diag(e^{-i \phi^{\alpha S}_{1}},\,\dots\,,e^{-i \phi^{\alpha S}_{i}}\,,\dots\,,
e^{-i \phi^{\alpha S}_{n}})\,,\nonumber\\
F_{\alpha T} &=& diag(e^{-i \phi^{\alpha T}_{1}},\,\dots\,,
e^{-i \phi^{\alpha T}_{i}}\,,\dots\,, e^{-i \phi^{\alpha T}_{n}})\,.
\end{eqnarray}

In the case that the mass matrix is Hermitian $T^{\alpha}$ can be replaced by
$ S^{\alpha}$, but only up to phases originating in the phases of the two basis,
the left handed one and the right handed one, since they remain independent.

The non Hermitian mass matrices can be diagonalized
in two ways, that is either one diagonalizes
$M^{\alpha }M^{\alpha \,\dagger}$ or $M^{\alpha \dagger} M^{\alpha }$
\begin{eqnarray}
\label{diagMM}
(S^{\alpha \dagger} M^{\alpha} T^{\alpha})          (S^{\alpha \dagger} M^{\alpha } T^{\alpha})^{\dagger}&=&
S^{\alpha \dagger} M^{\alpha } M^{\alpha \,\dagger} S^{\alpha} = {\bf M}^{\alpha  2}_{d S}\,, \nonumber\\
(S^{\alpha \dagger} M^{\alpha} T^{\alpha})^{\dagger}(S^{\alpha \dagger} M^{\alpha } T^{\alpha})&=&
T^{\alpha \dagger} M^{\alpha\, \dagger} M^{\alpha } T^{\alpha} = {\bf M}^{\alpha  2}_{d T}\,, \nonumber\\
{\bf M}^{\alpha\, \dagger }_{d S}&=& {\bf M}^{\alpha }_{d S}\, ,
\quad {\bf M}^{\alpha\, \dagger }_{d T}= {\bf M}^{\alpha}_{d T}\,.
\end{eqnarray}
The proof that ${\bf M}^{\alpha }_{d S}={\bf M}^{\alpha }_{d T}$ proceeds as
Let us define two Hermitian ($H^{\alpha}_{S}\,$,
$H^{\alpha}_{T}$) and two unitary matrices ($U^{\alpha}_{S}\,$, $H^{\alpha}_{T}$)
\begin{eqnarray}
\label{proof1}
H^{\alpha}_{S} &=& S^{\alpha} {\bf M}^{\alpha }_{d S} S^{\alpha \, \dagger}\,, \quad \quad
H^{\alpha}_{T}  =  T^{\alpha} {\bf M}^{\alpha \dagger }_{d T} T^{\alpha \, \dagger}\,,\nonumber\\
U^{\alpha}_{S} &=& H^{\alpha -1}_{S} M^{\alpha } \,, \quad \quad U^{\alpha}_{T} = H^{\alpha -1}_{T}
M^{\alpha \,\dagger } \,,
\end{eqnarray}
It is easy to show  that $H^{\alpha\, \dagger}_{S}= H^{\alpha}_{S}$,
$H^{\alpha\, \dagger}_{T}= H^{\alpha}_{T}$, $U^{\alpha}_{S} \,U^{\alpha\, \dagger}_{S}=I$ and
$U^{\alpha}_{T} \,U^{\alpha\, \dagger}_{T}=I$.
Then it follows
\begin{eqnarray}
\label{proof2}
S^{\alpha \dagger}\, H^{\alpha}_{S} \,S^{\alpha} &=& {\bf M}^{\alpha }_{d S}= {\bf M}^{\alpha \,\dagger}_{d S}=
S^{\alpha \dagger}\,M^{\alpha }\,U^{\alpha \,-1}_{S} \,S^{\alpha}= S^{\alpha \dagger}\,M^{\alpha }\, T^{\alpha} \,,
\nonumber\\
T^{\alpha \dagger}\, H^{\alpha}_{T} \,T^{\alpha} &=& {\bf M}^{\alpha }_{d T}= {\bf M}^{\alpha \,\dagger}_{d T}=
T^{\alpha \dagger}\,M^{\alpha \, \dagger}\,U^{\alpha \,-1}_{T} \,T^{\alpha}= T^{\alpha \dagger}\,
M^{\alpha \dagger}\, S^{\alpha} \,,
\end{eqnarray}
where we recognized $U^{\alpha \,-1}_{S} \,S^{\alpha}= T^{\alpha}$ and $U^{\alpha \,-1}_{T} \,T^{\alpha}=S^{\alpha}$.
Taking into account Eq.~(\ref{diagnonher}) the starting basis can be chosen so, that all diagonal
masses are real and positive.

\section{Statements and proofs of statements~[14]}
\label{proofs}
% To be read again

There are two kinds of the Clifford algebra objects, $\gamma^a$'s and
$\tilde{\gamma}^a$'s, both expressible with the Grassmann algebra objects,
$\theta^a$'s and their derivatives $\frac{\partial}{\partial \theta_a}$'s.
In Grassmann $d$-dimensional space there are $d$ anticommuting operators $\theta^{a}$,
$\{\theta^{a}, \theta^{b}\}_{+}=0$, $a=(0,1,2,3,5,..,d)$, and $d$ anticommuting derivatives
with respect to $\theta^{a}$, $\frac{\partial}{\partial \theta_{a}}$,
$\{\frac{\partial}{\partial \theta_{a}}, \frac{\partial}{\partial \theta_{b}}\}_{+} =0$, offering
together $2\cdot2^d$ operators, the half of which are superposition of products of
$\theta^{a}$ and another half corresponding superposition of
$\frac{\partial}{\partial \theta_{a}}$.
\begin{eqnarray}
\label{thetaderanti}
\{\theta^{a}, \theta^{b}\}_{+}=0\,, \, && \,
\{\frac{\partial}{\partial \theta_{a}}, \frac{\partial}{\partial \theta_{b}}\}_{+} =0\,,
\nonumber\\
\{\theta_{a},\frac{\partial}{\partial \theta_{b}}\}_{+} &=&\delta_{ab}\,, (a,b)=(0,1,2,3,5,\cdots,d)\,.
\end{eqnarray}
\begin{eqnarray}
(\theta^{a})^{\dagger} &=& \eta^{a a} \frac{\partial}{\partial \theta_{a}}\,,\quad
{\rm it \, follows}\,,\quad
(\frac{\partial}{\partial \theta_{a}})^{\dagger}= \eta^{a a} \theta^{a}\,.
\label{thetaderher}
\end{eqnarray}
The identity is the self adjoint member.
The signature $\eta^{a b}=diag\{1,-1,-1,\cdots,-1\}$ is assumed.

One can define  %$2 \cdot 2^{d}$
new operators, expressed with $\theta^{a}$'s  and
$\frac{\partial}{\partial \theta_{a}}$'s
\begin{eqnarray}
\label{cliffthetaapp}
\gamma^{a} &=& (\theta^{a} + \frac{\partial}{\partial \theta_{a}})\,, \quad
\tilde{\gamma}^{a} =i \,(\theta^{a} - \frac{\partial}{\partial \theta_{a}})\,,\nonumber\\
\theta^{a} &=&\frac{1}{2} \,(\gamma^{a} - i \tilde{\gamma}^{a})\,, \quad
\frac{\partial}{\partial \theta_{a}}= \frac{1}{2} \,(\gamma^{a} + i \tilde{\gamma}^{a})\,.
%(\gamma^{a})^{\dagger} &=&\eta^{aa}\, \gamma^{a}\,, \quad
%(\tilde{\gamma}^{a})^{\dagger} = \eta^{aa} \,\tilde{\gamma}^{a}\,,
\end{eqnarray}
One  can make $2^d$  products of superpositions of $\gamma^a$'s and  $2^{d}$  products of superposition of $\tilde{\gamma}^{a}$'s, all together  $2\cdot 2^{d}$

{\bf  Statement 1.} $\gamma^a$'s  and $\tilde{\gamma}^{a}$'s define two independent Clfford algebras.

To prove this statement one only needs to take into account Eqs.~(\ref{thetaderher}, \ref{clifftheta}).
\begin{eqnarray}
\label{gammatildeantiherapp}
\{\gamma^{a}, \gamma^{b}\}_{+}&=&2 \eta^{a b}= \{\tilde{\gamma}^{a},
\tilde{\gamma}^{b}\}_{+}\,, \nonumber\\
\{\gamma^{a}, \tilde{\gamma}^{b}\}_{+}&=&0\,,\quad
(a,b)=(0,1,2,3,5,\cdots,d)\,, \nonumber\\
(\gamma^{a})^{\dagger} &=& \eta^{aa}\, \gamma^{a}\, , \quad
(\tilde{\gamma}^{a})^{\dagger} =  \eta^{a a}\, \tilde{\gamma}^{a}\,.%\nonumber\\
% S^{ab} &=& \frac{i}{2}\, (\gamma^a \gamma^b - \gamma^b \gamma^a)\,, \quad
% \tilde{S}^{ab} = \frac{i}{2}\, (\tilde{\gamma}^a \tilde{\gamma}^b - \tilde{\gamma}^b
% \tilde{\gamma}^a)\,,
\end{eqnarray}
Eq.~(\ref{gammatildeantiherapp}) demonstrates that these two Clifford algebra
objects obviously define two independent ''basis vectors''. Either $\gamma^a$'s or
$\tilde{\gamma}^a$'s are, up to a sign, self adjoint operators.

The generators of the Lorentz transformations in the Grassmann algebra space are
defined as follows
\begin{eqnarray}
{\cal {\bf S}}^{a b} &=& i \, (\theta^{a} \frac{\partial}{\partial \theta_{b}} - \theta^{b}
\frac{\partial}{\partial \theta_{a}})\,, \quad  ({\cal {\bf S}}^{a b})^{\dagger} = \eta^{a a}
\eta^{b b} {\cal {\bf S}}^{a b}\, .
\label{thetasabapp}
\end{eqnarray}

{\bf Statement 2:}
The sum of the generators of the Lorentz transformations in
each of the two Clifford algebra spaces,
$S^{ab}=i \frac{1}{4} (\gamma^{a} \gamma^{b}-
\gamma^{b}\gamma^{a})$ and
$\tilde{S}^{ab}=i \frac{1}{4}
(\tilde{\gamma}^{a}\tilde{\gamma}^{b}- \tilde{\gamma}^{b}\tilde{\gamma}^{a})$,
respectively, are equal to the generators of the Lorentz transformation in the
Grassmann algebra space $\bf{\cal S}^{a b}=S^{ab} + \tilde{S}^{ab}$.

To prove this statement one only has to express in the sum $S^{ab} + \tilde{S}^{ab}$ first $S^{ab}$ with $\gamma^a$ and $\gamma^b$ and $\tilde{S}^{ab}$ with
$\tilde{\gamma}^a$ and $\tilde{\gamma}^b$, and then $\gamma^{a}$ and
$\tilde{\gamma}^{a}$ with  $\theta^{a}$ and $\frac{\partial}{\partial \theta_{a}}$,
using Eq.~(\ref{clifftheta}).

One obtains  that
$S^{ab}=i \frac{1}{2}(\frac{\partial}{\partial \theta_{a}}\theta^b +
\theta^a \frac{\partial}{\partial \theta_{b}})$ and $\tilde{S}^{ab}=i \frac{1}{2}(
\theta^a \frac{\partial}{\partial \theta_{b}} +
\frac{\partial}{\partial \theta_{a}}\theta^b)$, leading to $S^{ab} + \tilde{S}^{ab}=
{\cal{\bf S}}^{ab}$.

We conclude
\begin{eqnarray}
S^{ab} &=&\frac{i}{4}(\gamma^a \gamma^b - \gamma^b \gamma^a)\,,
\quad
\tilde{S}^{ab} =\frac{i}{4}(\tilde{\gamma}^a \tilde{\gamma}^b - \tilde{\gamma}^b
\tilde{\gamma}^a)\, , \nonumber\\
{\cal {\bf S}}^{ab} &=&S^{ab} + \tilde{S}^{ab}\,, \quad
\{S^{ab}, \tilde{S}^{ab}\}_{-}=0\,,\nonumber\\
\{ S^{ab}, \gamma^c\}_{-}&=& i (\eta^{bc} \gamma^a - \eta^{ac} \gamma^b)\,, \nonumber\\%\quad
\{ \tilde{S}^{ab}, \tilde{\gamma}^c \}_{-}&=& i (\eta^{bc} \tilde{\gamma}^a -
\eta^{ac }\tilde{\gamma}^b)\,,\nonumber\\
\{ S^{ab}, \tilde{\gamma}^c\}_{-}&=&0\,,\quad \{\tilde{S}^{ab}, \gamma^c\}_{-}=0\,.
\label{sabtildesabapp}
\end{eqnarray}

{\bf Statement 2a:}
The eigenvectors of the operators ${\cal {\bf S}}^{ab}=
i \, (\theta^{a} \frac{\partial}{\partial \theta_{b}} - \theta^{b}
\frac{\partial}{\partial \theta_{a}})\,$ can be  written as follows
\begin{eqnarray}
{\cal {\bf S}}^{ab} \,\frac{1}{\sqrt{2}}\, (\theta^a + \frac{\eta^{aa}}{i k} \theta^b) &=&
k\,\frac{1}{\sqrt{2}} (\theta^a + \frac{\eta^{aa}}{ik} \theta^b) \,, \nonumber\\
{\cal {\bf S}}^{ab} \,\frac{1}{\sqrt{2}}\, (1+ \frac{i}{k}  \theta^a \theta^b) &=&0\,,
% \nonumber\\ &{\rm or} &\;  \nonumber\\
%{\cal {\bf S}}^{ab} \,\frac{1}{\sqrt{2}}\,  \frac{i}{k}  \theta^a \theta^b &=&0\,,
\label{eigengrasscartanapp}
\end{eqnarray}
while the corresponding eigenvectors of $S^{ab}$ and  $\tilde{S}^{ab}$ in each
of the two spaces are
\begin{eqnarray}
S^{ab} \frac{1}{2} (\gamma^a + \frac{\eta^{aa}}{ik} \gamma^b) &=& \frac{k}{2}  \,
\frac{1}{2} (\gamma^a + \frac{\eta^{aa}}{ik} \gamma^b)\,,\nonumber\\
S^{ab} \frac{1}{2} (1 +  \frac{i}{k}  \gamma^a \gamma^b) &=&  \frac{k}{2}  \,
\frac{1}{2} (1 +  \frac{i}{k}  \gamma^a \gamma^b)\,,\nonumber\\
\tilde{S}^{ab} \frac{1}{2} (\tilde{\gamma}^a + \frac{\eta^{aa}}{ik} \tilde{\gamma}^b) &=&
\frac{k}{2}  \,\frac{1}{2} (\tilde{\gamma}^a + \frac{\eta^{aa}}{ik} \tilde{\gamma}^b)\,,
\nonumber\\
\tilde{S}^{ab} \frac{1}{2} (1 +  \frac{i}{k}  \tilde{\gamma}^a \tilde{\gamma}^b) &=&
\frac{k}{2}  \, \frac{1}{2} (1 +  \frac{i}{k} \tilde{\gamma}^a \tilde{\gamma}^b)\,.
\label{eigencliffcartanapp}
\end{eqnarray}
with $k^2 = \eta^{aa} \eta^{bb}$.

The Eq.~(\ref{eigengrasscartan}) can be proven by applying ${\cal {\bf S}}^{ab} $
on $\frac{1}{\sqrt{2}}\, (\theta^a + \frac{\eta^{aa}}{i k} \theta^b)$, leading to

$i \frac{1}{\sqrt{2}} (- \theta^b \eta^{aa} +
\frac{\eta^{aa}\eta^{bb}}{ik} \theta^a)= \frac{ik^2}{ik \sqrt{2}} (\theta^a-
\frac{\eta^{aa} ik}{k^2} \theta^b)$.

The application of ${\cal {\bf S}}^{ab} $ on either
a constant or on a constant$\cdot \theta^a \theta^b$ gives $0$. In all these cases
it is assumed that $a\ne b$.

The proof of Eq.(\ref{eigencliffcartanapp}) goes similarly, again $a\ne b$ is assumed:

$\frac{i}{2}\gamma^a \gamma^b \frac{1}{2}  (\gamma^a +
\frac{\eta^{aa}}{ik} \gamma^b)=\frac{i}{2} \frac{1}{2}  (-\eta^{aa}\gamma^b +
\frac{\eta^{aa}\eta^{bb}}{ik} \gamma^a) = \frac{k}{2}\frac{1}{2}(\gamma^a-
\eta^{aa}\frac{i}{k} \gamma^b)$.

$\frac{i}{2}\gamma^a \gamma^b \frac{1}{2}  ( 1+\frac{i}{k}\gamma^a \gamma^b)
=\frac{i}{2} \frac{1}{2}(\gamma^a \gamma^b -\frac{i}{k}\eta^{aa} \eta^{bb})=
\frac{k}{2} \frac{1}{2}(1 +  \frac{i}{k} \gamma^a \gamma^b)$.

Replacing $S^{ab}$ with $\tilde{S}^{ab}$, and $\gamma^a$'s with
$\tilde{\gamma}^a$'s goes through the same steps.

{\bf Statement 2b:}
The members of any irreducible representation of $S^{ab}$ follow
from the starting one by the application of $S^{cd}$, which do not belong
to the Cartan subalgebra of the Lorentz algebra.

The proof follows if we apply $\gamma^a$ on nilpotents and projectors, since
$S^{ab}=\frac{i}{2}\gamma^a \gamma^b$.

\begin{eqnarray}
&&\gamma^a \stackrel{ab}{(k)}= \eta^{aa}\stackrel{ab}{[-k]},\; \quad
\gamma^b \stackrel{ab}{(k)}= -ik \stackrel{ab}{[-k]}, \; \quad
\gamma^a \stackrel{ab}{[k]}= \stackrel{ab}{(-k)},\;\quad
\gamma^b \stackrel{ab}{[k]}= -ik \eta^{aa} \stackrel{ab}{(-k)}\,.
%&&\tilde{\gamma^a} \stackrel{ab}{(k)} = - i\eta^{aa}\stackrel{ab}{[k]},\;\quad
%\tilde{\gamma^b} \stackrel{ab}{(k)} =  - k \stackrel{ab}{[k]}, \;\nonumber\\
%&&\tilde{\gamma^a} \stackrel{ab}{[k]} =  \;\;i\stackrel{ab}{(k)},\; \quad
%\tilde{\gamma^b} \stackrel{ab}{[k]} =  -k \eta^{aa} \stackrel{ab}{(k)}\,.
\label{snmb:gammatildegammaapp}
\end{eqnarray}
Correspondingly,  any transformation on  ''basis vectors'' of the kind, which
do not change sign as is required in Eq.~(\ref{snmb:gammatildegammaapp}),
lead to another irreducible representations.

{\bf Statement 3:}
Postulating that  $\tilde{\gamma}^{a}$'s operate on
$\gamma^a$'s as follows~%
[67, 9, 19, 21, 22]
\begin{eqnarray}
\tilde{\gamma}^a B &=&(-)^B\, i \, B \gamma^a\,,\nonumber
\end{eqnarray}
with $(-)^B = -1$, if $B$ is (a function of) an odd product of $\gamma^a$'s,
otherwise $(-)^B = 1$~[67], the reduction of the Clifford space and
correspondingly also the reduction of the Grassmann space follows.

Eq.~(\ref{tildegammareduced}) requires
\begin{eqnarray}
\label{partialthetazeroev}
&&[\tilde{\gamma}^a (a_0 + a_{bc}\gamma^b \gamma^c + a_{bcde}\gamma^b
\gamma^c \gamma^d \gamma^e +\cdots) =  i  (a_0 + a_{bc}\gamma^b \gamma^c + a_{bcde}\gamma^b \gamma^c \gamma^d \gamma^e +\cdots) \gamma^a ]
|\psi_{oc}>\,, \nonumber\\
&&[\tilde{\gamma}^a (a_{b}\gamma^b + a_{bcd}\gamma^b
\gamma^c \gamma^d +\cdots) = - i  (a_{b}\gamma^b + a_{bcd}\gamma^b
\gamma^c \gamma^d + \cdots) \gamma^a ]
|\psi_{oc}>\,, \nonumber
\end{eqnarray}

To prove {\it Statement 3.} let us evaluate what does Eq.~(\ref{tildegammareduced})
require when we use Eq.(\ref{clifftheta})on $|\psi_{oc}>$: $ \gamma^{a} = (\theta^{a} + \frac{\partial}{\partial \theta_{a}})$ and $\tilde{\gamma}^{a} =i \,(\theta^{a} - \frac{\partial}{\partial \theta_{a}})$, with $|\psi_{oc}>$ expressed as well with $\theta^a$'s and
$ \frac{\partial}{\partial \theta_{a}} $.

Let us point out that $|\psi_{oc}>$, expressed in terms of $\theta^a$'s and
$ \frac{\partial}{\partial \theta_{a}}$'s is an even function of  $\theta^a$'s,
$|\psi_{oc}>=(1 + \theta^a\theta^b +\cdots)$, while
$ \frac{\partial}{\partial \theta_{a}} $, applying on identity, gives zero.
% Show on one example! d=(5+1)

The proof is needed for any even and any odd summand of $B$, appearing in  Eq.~(\ref{tildegammareduced}) and for an arbitrary $|\psi_{oc}>$.

% * 31.12.2020 at 17:30\\

Let us start with $[\tilde{\gamma}^a a_0=i a_0 \gamma^a ] |\psi_{oc}>$,
with $|\psi_{oc}>=(1 + \theta^a\theta^b +\cdots)$ and $a_{0}$ an arbitrary
constant. This relation requires that
$[i(\theta^a- \frac{\partial}{\partial \theta_{a}})=
i(\theta^a +\frac{\partial}{\partial \theta_{a}})] |\psi_{oc}>$, leading to $- 2i \frac{\partial}{\partial \theta_{a}}|\psi_{oc}>
\Rightarrow 0, \; \forall \frac{\partial}{\partial \theta_{a}}$.
This last relation can only be true if $\frac{\partial}{\partial \theta_{a}} |\psi_{oc}>
\Rightarrow 0,\;\forall \frac{\partial}{\partial \theta_{a}}$.

Evaluating $[\tilde{\gamma}^a a_{bc}\gamma^b \gamma^c=
i a_{bc}\gamma^b \gamma^c \gamma^a] |\psi_{oc}>$ we end up again
with the requirement $\frac{\partial}{\partial \theta_{a}} |\psi_{oc}>
\Rightarrow 0,\;
\forall \frac{\partial}{\partial \theta_{a}}$.
Applying  $\tilde{\gamma}^a$ on any even products of $\gamma^a$'s  we end
up with the same requirement $\frac{\partial}{\partial \theta_{a}} |\psi_{oc}>
\Rightarrow 0,\; \forall \frac{\partial}{\partial \theta_{a}}$.

Application of $\tilde{\gamma}^a$ on any odd products of $\gamma^a$'s,
while $|\psi_{oc}>=(1 + \theta^a\theta^b +\cdots)$,
% CONTINUE and add proofs of \footnote [55.54,56]
%$[\frac{\partial}{\partial \theta_{a}}$
\begin{eqnarray}
\label{partialthetazerood}
&&[\tilde{\gamma}^a (a_{b}\gamma^b + a_{bcd}\gamma^b
\gamma^c \gamma^d +\cdots) = - i  (a_{b}\gamma^b + a_{bcd}\gamma^b
\gamma^c \gamma^d + \cdots) \gamma^a ]
\end{eqnarray}
it follows again,  after expressing $\tilde{\gamma}^a=
i(\theta^a - \frac{\partial}{\partial \theta_{a}})$ and $\gamma^a=
(\theta^a + \frac{\partial}{\partial \theta_{a}})$ into Eq.~\ref{partialthetazerood},
% (in the relation, for example, $[\tilde{\gamma}^a  a_{b}\gamma^b = - i a_{b}\gamma^b \gamma^a ] |\psi_{oc}>$),
that $\frac{\partial}{\partial \theta_{a}} |\psi_{oc}>\Rightarrow 0 $ is the only solution,
leading to .
% The same result follows for the application of $\tilde{\gamma}^a$'s on an arbitrary  % odd product of $\gamma^a$'s. Then it follows
\begin{eqnarray}
\theta^a \Rightarrow \gamma^a\,, %, \quad \frac{\partial}{\partial \theta_{a}} =0\,.
\nonumber
\label{tildegammareduced1}
\end{eqnarray}
which does not mean that $\theta^a$ is equal to $\gamma^a$ but rather that the whole
Grassmann algebra reduces to only one of the two Clifford algebras, the one, in which
the ''basis vectors''  are superposition of products of (odd when describing fermions)
number of $\gamma^a$'s. It also does not mean that $\theta^a$'s are equal to
$i\tilde{\gamma}^a$'s, since the application of $\tilde{\gamma}^a$'s depend on
properties of $B(\gamma^a)$, on which $\tilde{\gamma}^a$'s apply.

{\bf Statement 3a:}
The relations of  Eq.~(\ref{gammatildeantiher}) remain valid also after the reduction of the   Clifford space.

% One can  check the relations presented in  Eq.~(\ref{gammatildeantiher}) that
% all  the relations of this equations remain valid also after the reduction of the
% Clifford space~\footnote{
Let us check  Eq.~(\ref{gammatildeantiher}):
$\{ \tilde{\gamma}^{a}, \tilde{\gamma}^{b}\}_{+}= 2\eta^{ab}=$
$\tilde{\gamma}^{a} \tilde{\gamma}^{b}+\tilde{\gamma}^{b}
\tilde{\gamma}^{a}=$ $ \tilde{\gamma}^{a} i\gamma^b +\tilde{\gamma}^{b} i \gamma^a=$
$ i \gamma^b (-i)\gamma^a + i\gamma^a(-i)\gamma^b= 2\eta^{ab} $.
$\{ \tilde{\gamma}^{a}, \gamma^b\}_{+}= 0=$
$\tilde{\gamma}^{a} \gamma^b+\gamma^b \tilde{\gamma}^{a}=$
$ \gamma^b (-i)\gamma^a+  \gamma^b i \gamma^a=0$.
For a particular case one has $\{ \tilde{\gamma}^{a}, \gamma^a\}_{+}= 0=$
$ \tilde{\gamma}^{a} \gamma^a + \gamma^a \tilde{\gamma}^a= $
$\gamma^a (-i )\gamma^a +\gamma^a i \gamma^a=0$.

The application of $\tilde{\gamma}^a$  obviously depends on the space on which
it applies, Eq.~(\ref{tildegammareduced}) namely requires:
$[\tilde{\gamma}^a (a_0 + a_b\gamma^b +
a_{bc} \gamma^b \gamma^c+ \cdots ) =
( i a_0 \gamma^a + (-i) a_b  \gamma^b \gamma^a +
i a_{bc} \gamma^b \gamma^c  \gamma^a +\cdots)] |\psi_{oc>}$.
Statement 3b is proved.

%* Correct above!!

% The Hermitian conjugated part of the space in the Grassmann case is "freezed out"
% together with the "vector" space of $\tilde{\gamma}^a$'s.

{\bf Statement 3b:}
Taking into account in Eq.~(\ref{tildegammareduced}) required application of $\tilde{\gamma}^a$'s on the Clifford space of $\gamma^a$'s (causing the reduction of the Clifford space and at the same time as well
the reduction of the Grassmann space),
it follows that the eigenvalues of  $\tilde{S}^{ab}$ on the eigenvectors of $S^{ab}$
agree with the eigenvalues of $S^{ab}$ on nilpotents, while  the eigenvalues of
$\tilde{S}^{ab}$ and  $S^{ab}$ on projectors, which are eigenvectors of $S^{ab}$, have opposite sign.

Let us check Eq.(\ref{signature0}).

$\tilde{S}^{ab}\frac{1}{2}(\gamma^a + \frac{\eta^{aa}}{ik}\gamma^b)=
\frac{i}{2} \tilde{\gamma}^a \tilde{\gamma}^b \frac{1}{2}(\gamma^a +
\frac{\eta^{aa}}{ik}\gamma^b) = \frac{i}{2} \frac{1}{2}(\gamma^a +
\frac{\eta^{aa}}{ik}\gamma^b) \gamma^b \gamma^a= $
$\frac{i}{2} \frac{1}{2}(-\eta^{aa} \gamma^b +\frac{\eta^{aa} \eta^{bb}}{ik}
\gamma^a)$ $=\frac{k}{2} \frac{1}{2}(\gamma^a +
\frac{\eta^{aa}}{ik}\gamma^b)$,

$\tilde{S}^{ab}\frac{1}{2}( 1+  \frac{i}{k} \gamma^a\gamma^b)=
\frac{i}{2} \frac{1}{2}(1 + \frac{i}{k}\gamma^a\gamma^b)
\gamma^b \gamma^a= $ $\frac{i}{2} \frac{1}{2}(- \gamma^a\gamma^b +
\frac{i}{k} \eta^{aa} \eta^{bb}) =- \frac{k}{2}\frac{1}{2} ( 1+  \frac{i}{k}
\gamma^a\gamma^b)$,
where it is taken into account that $k^2=\eta^{aa} \eta^{bb}$.\\
This proves  {\it Statement 3b}.

{\bf Statement 4:}
The algebraic product of
$\hat{b}^m_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}$ is  the same for all
$m$ of a particular irreducible representation $f$.

To prove this we take into account that $\hat{b}^{m \dagger}_{f}$  follows
from $\hat{b}^{m -1\dagger}_{f}$ by the application of a particular  $2i S^{eg}$.
Then $\hat{b}^{m}_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}  =$
$\hat{b}^{m-1}_{f} (2\,S^{eg})^{\dagger}{}{*_{A}} (2\,S^{eg})
\hat{b}^{m-1 \dagger}_{f}=$
$\hat{b}^{m-1}_{f}{}{*_{A}}\hat{b}^{m-1 \dagger}_{f}$, due to the relation
$(-2i S^{ab})^{\dagger} (-2i S^{ab})=1$). Repeating this procedure for each $m$
proves the statement.

{\bf Statement 5:}
Each irreducible representation has its own algebraic product
$\hat{b}^m_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}$.

We pay attention to the
Clifford odd representations, but the proof is as well valid for  the Clifford even

To prove this statement let us start with $d=2(2n+1)$ with the irreducible
representation $f$ and the member  $m$ equal to $\hat{b}^{m \dagger}_{f}=
\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1\, d}{(+)}$. To obtain
the $2^{\frac{d}{2}- 1}-1$  rest irreducible representations we must transform
each of possible pairs  $\stackrel{ab}{(k)}\stackrel{cd}{(k)}$ into
$\stackrel{ab}{[k]}\stackrel{cd}{[k]}$. Let us start with the first two. One obtains
$\hat{b}^{m \dagger}_{f'}=
\stackrel{03}{[+i]}\stackrel{12}{[+]}\cdots \stackrel{d-1\, d}{(+)}$.
The two algebraic products,  $\hat{b}^m_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}
=\stackrel{03}{[-i]}\stackrel{12}{[-]} \stackrel{56}{[-]}\cdots \stackrel{d-1\, d}{[-]}$ and $\hat{b}^m_{f'}{}{*_{A}}\hat{b}^{m \dagger}_{f'}
\stackrel{56}{[-]}\cdots \stackrel{d-1\, d}{[-]}$,
distinguish in the first two projectors. When replacing a pair by a pair in
$\hat{b}^{m \dagger}_{f}$, we end up with $2^{\frac{d}{2}- 1}$ different
$\hat{b}^m_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}$, differing in all possible
pairs $\stackrel{ab}{[+]}\stackrel{a'b'}{[+]}$ replacing $\stackrel{ab}{[-]}\stackrel{a'b'}{[-]}$.

For $d=4n$  and the Clifford odd representations we must start with
$\hat{b}^{m \dagger}_{f}=
\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1\, d}{[+]}$,
and then repeat all steps. We shall again obtain $2^{\frac{d}{2}- 1}$  different
$\hat{b}^m_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}$.

{\bf Statement 5a:}
There are  $2^{\frac{d}{2}- 1}$  different algebraic products
$\hat{b}^m_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}$.

Since due to {\it Statement 4.}  all the members of a particular  irreducible
representation have the same algebraic product
$\hat{b}^m_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}$, we can conclude that there
are  $2^{\frac{d}{2}- 1}$  different algebraic products
$\hat{b}^m_{f}{}{*_{A}}\hat{b}^{m \dagger}_{f}$.
%* Mention how do we obtain different irreducible representations.

{\bf Statement 5b:}
Each creation operator $\hat{b}^{m \dagger}_{f}$ gives
nonzero contribution when applied on $|\psi_{oc}>$.

There is one summand in $|\psi_{oc}>$, namely, $\hat{b}^{m}_{f}
\hat{b}^{m \dagger}_{f}=\hat{b}^{m'}_{f} \hat{b}^{m' \dagger}_{f}$,
$\forall m'$, on which $\hat{b}^{m \dagger}_{f}$ gives a nonzero contribution:
$\sum_{f'=1}^{2^{\frac{d}{2}-1}} \hat{b}^{m \dagger}_{f}\,$
$\hat{b}^{m}_{f'}\, \hat{b}^{m \dagger}_{f'}=$
$\sum_{f'=1}^{2^{\frac{d}{2}-1}} \,
\delta^{f f'} \hat{b}^{m \dagger}_{f'} $.

{\bf Statement 5c:}
In odd representations the algebraic product of any two
annihilation operators  $\hat{b}^{m}_{f}{}_{*_A} \hat{b}^{m'}_{f'}$ gives
zero, as also the algebraic product of any two creation operators
$\hat{b}^{m \dagger}_{f}{}_{*_A} \hat{b}^{m' \dagger}_{f'}$ gives
zero. Correspondingly the application of the annihilation operators on
$|\psi_{oc}>$ gives  zero contribution.

All annihilation operators are ''orthogonal'', as also all the creation operators are:
$\hat{b}^{m}_{f}{}_{*_A} \hat{b}^{m'}_{f'}=0$,
$\hat{b}^{m \dagger}_{f}{}_{*_A} \hat{b}^{m' \dagger}_{f'}=0$.
Within the same irreducible representation at least one nilpotent of the two creation
operators or  of the two annihilation  operators are  the same.
Among different irreducible representations of each kind separately, either
one nilpotent is the same in both operators appearing in the product,  or
$\stackrel{ab}{(k)}$ multiplies $\stackrel{ab}{[k ]}$ or
$\stackrel{ab}{[-k]}$ multiplies $\stackrel{ab}{(k)}$ or
$\stackrel{ab}{[k]}$ multiplies $\stackrel{ab}{[-k ]}$, since one irreducible representation differs from the other in a pair $\stackrel{ab}{(k)}\stackrel{ef}{(k')}$
going to   $\stackrel{ab}{[k]}\stackrel{ef}{[k']}$ or
$\stackrel{ab}{(k)}\stackrel{ef}{[k']}$ going to  $\stackrel{ab}{[k]}
\stackrel{ef}{(k')}$ or $\stackrel{ab}{[k]}\stackrel{ef}{(k')}$ going
to $\stackrel{ab}{(k)}\stackrel{ef}{[k']}$.

%* Check again! Explain better!!

%* 01.01.2021.

{\bf Statement 6:}
The operator $\tilde{S}^{cd}$, which does not belong to the
Cartan subalgebra of  Eq,~(\ref{cartangrasscliff}), generates after the reduction
of the Clifford space  a new irreducible representation, carrying different
family quantum number.

%, determined by the eigenvalues of the members of ..

The proof of {\it Statement 5.} %and {\it Statement 4.}
contains also the proof for  {\it Statement 6.}.   All the members of one irreducible representation are reachable by  the application of $S^{ab}$'s. Let us start in
$d=2(2n+1)$ with the Clifford odd representation containing the member
$\hat{b}^{m \dagger}_{f}=
\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{b d}{(+)} \cdots
\stackrel{b' d'}{(+)}\cdots\stackrel{d-1\, d}{(+)}$ (with the family quantum numbers
$(\frac{i}{2}, \frac{1}{2},\cdots, \frac{1}{2}, \cdots, \frac{1}{2}, \cdots,
\frac{1}{2} )$, determined by $\tilde{S}^{ab}$'s from Eq.~(\ref{cartangrasscliff})).
Operator $\tilde{S}^{b b'}$  transforms   $\hat{b}^{m \dagger}_{f}$
into $\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{b d}{[+]} \cdots
\stackrel{b' d'}{[+]}\cdots \stackrel{d-1\, d}{(+)}$. This  new creation operator
belongs to new  irreducible representation (with the family quantum numbers
$(\frac{i}{2}, \frac{1}{2},\cdots,- \frac{1}{2}, \cdots, -\frac{1}{2}, \cdots,
\frac{1}{2})$), since it is not reachable by $S^{ab}$
(which generate all the rest members of the same irreducible representation).
Transforming all pairs into the new ones, one obtains $2^{\frac{d}{2}-1}$

For $d=4n$  we start for odd representations by $\hat{b}^{m \dagger}_{f}=
\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{b d}{(+)} \cdots
\stackrel{b' d'}{(+)}\cdots\stackrel{d-1\, d}{[+]}$ and repeat the above
procedure (by taking into account that $\tilde{S}^{b b'}$  transforms
$\stackrel{b d}{(+)} \cdots \stackrel{b' d'}{[+]}$ into $\stackrel{b d}{[+]} \cdots
\stackrel{b' d'}{(+)}$).

Similarly we can find all the families of the Clifford even representations, if
taking into account Eq.~(\ref{tildegammareduced})

%Due to {\it Statement 4.}

% *** Make it! See {\it Statement 5a.} and put them together.\\

{\bf Statement 7:}
Creation operators $\hat{b}^{m \dagger}_{f}$ and their
Hermitian conjugated partners annihilation operators, appearing in
Eq.~(\ref{almostDirac}), have the properties
\begin{eqnarray}
\label{almostDirac1}
\hat{b}^{m}_{f} {}_{*_{A}}|\psi_{oc}>&=& 0\, \cdot|\psi_{oc}>\,,\nonumber\\
\hat{b}^{m \dagger}_{f}{}_{*_{A}}|\psi_{oc}>&=&  |\psi^m_{f}>\,,\nonumber\\
\{\hat{b}^{m}_{f}, \hat{b}^{m'}_{f'}\}_{*_{A}+}|\psi_{oc}>&=&
0\,|\psi_{oc}>\,, \nonumber\\
\{\hat{b}^{m }_{f}, \hat{b}^{m' \dagger}_{f}\}_{*_{A}+}|\psi_{oc}>
\{\hat{b}^{m  \dagger}_{f}, \hat{b}^{m' \dagger}_{f'}\}_{*_{A}+}|\psi_{oc}>
%%<\psi_{oc}| \hat{b}^{m' \dagger}_{f'}\, *_{A}\,\hat{b}^{m \dagger}_{f}
%*_{A}|\psi_{oc}> &=& \delta^{m m'} \delta_{f f'}\,.
\end{eqnarray}
Let us prove this statement step by step:\\
a. The last line  of Eq.~(\ref{almostDirac1}) requires that $[\hat{b}^{m  \dagger}_{f}, \hat{b}^{m' \dagger}_{f'}=0] |\psi_{oc}>$. \\
a.i. The product of two equal creation operators is zero, since the product of two
nilpotents of the same kind, $(\stackrel{ab}{(k)})^2=0$, gives zero.\\
a.ii. All the creation operators can be obtained from the starting one by the
application $S^{cd}$.  Since at this application at least one nilpotent remains
the same, it follows that all creation operators within the same irreducible
representation are orthogonal
($\hat{b}^{m \dagger}_{f}{}_{*_A}\hat{b}^{m' \dagger}_{f}=0$),
by themselves and then also on $|\psi_{oc}>$. \\
a.iii. The creation operators of two different irreducible representations $f\ne f'$
can be obtained from the starting one by replacing two nilpotents
$\stackrel{ab}{(k)}{}_{*_A}\stackrel{ef}{(k')}$ by
a nilpotent $\stackrel{ab}{(k)}$ and a projector $\stackrel{ab}{[k]}$
by the projector $\stackrel{ab}{[k]}$ and the nilpotent $\stackrel{ab}{(k)}$,
or two projectors $\stackrel{ab}{[k]}$ $\stackrel{ab}{[k']}$ by two nilpotents.
Since at least one of the nilpotents remains the same, it follows that
all the same members of different irreducible representations are orthogonal,
($\hat{b}^{m \dagger}_{f}{}_{*_A}\hat{b}^{m \dagger}_{f'}=0$),
by themselves and then also on $|\psi_{oc}>$.\\
a.iv. The creation operators, belonging to two different irreducible representations
$(f, f')$ and to two different members $(m,m')$ have the property
($\hat{b}^{m \dagger}_{f}{}_{*_A}\hat{b}^{m' \dagger}_{f'}=0$)
(are  orthogonal), due to the way how they are created
($\hat{b}^{m \dagger}_{f}{}_{*_A}\hat{b}^{m' \dagger}_{f'}
=$ $ S^{m m'}\hat{b}^{m' \dagger}_{f}{}_{*_A}\hat{b}^{m' \dagger}_{f'}=0$,
as it is proven under iii.); Either the two have the same nilpotent or
there appear a product of two projectors of the same type with opposite $k$
($\stackrel{ab}{[k]}$, $\stackrel{ab}{[-k]}$), or it appears
($\stackrel{ab}{[k]}{}_{*_A}\stackrel{ef}{(-k)}=0$) or
($\stackrel{ab}{(k)}{}_{*_A}\stackrel{ef}{[k]}=0$), Eq.~(\ref{graficcliff}).\\
%*01.01.2021. at 19:40
b. The first line requires $\hat{b}^m_{f}{}_{*_A}|\psi_{oc}>=0$, which is
equivalent to requiring $\hat{b}^{m}_{f}{}_{*_A}\hat{b}^{m'}_{f'}{}_{*_A}
\hat{b}^{m' \dagger}_{f'}=0$, $\forall \,m' \, {\rm and}\, f'$, since the vacuum
state is equal to $|\psi_{oc}>= \sum_{f=1}^{2^{\frac{d}{2}-1}}
\hat{b}^m_{f}{}{*_{A}} \hat{b}^{m \dagger}_{f}$, independent of the
choice of $m$. The proofs from a.i.-a.iv.
% together with the last one
guarantee that $\hat{b}^m_{f}{}_{*_A}|\psi_{oc}>=0$, since
$\hat{b}^m_{f}=(\hat{b}^{m \dagger}_{f})^{\dagger}$.
\\
c. The second line requires that $\hat{b}^{m' \dagger}_{f'}{}_{*_A}|\psi_{oc}>
= |\psi^{m'}_{f'}>$.
Namely, $\hat{b}^{m' \dagger}_{f'}{}_{*_A}\hat{b}^{m}_{f'}{}_{*_A}
\hat{b}^{m\dagger}_{f'}$ $=\hat{b}^{m' \dagger}_{f'}$ $\forall m$,  what follows from the way how $|\psi_{oc}>$ is created, while
$\hat{b}^{m' \dagger}_{f'}{}_{*_A}\hat{b}^{m}_{f}{}_{*_A}
\hat{b}^{m \dagger}_{f}=0$, $\forall \, f\ne f'$,
since the application of  $\hat{b}^{m' \dagger}_{f'}{}_{*_A}$ on
$\hat{b}^{m}_{f}{}_{*_A}\hat{b}^{m\dagger}_{f'}$ gives zero due to the
orthogonality of the members of different irreducible  representations.\\
% Say this more elegantly!
d. The third line requires of Eq.~(\ref{almostDirac1}) requires that $[\hat{b}^{m}_{f}, \hat{b}^{m'}_{f'}=0] |\psi_{oc}>$.  Since $\hat{b}^{m'}_{f'}=
(\hat{b}^{m' \dagger}_{f'})^{\dagger}$, the proof is the same as in the case a..\\
e. The fourth line requires that $\{\hat{b}^{m }_{f}, \hat{b}^{m' \dagger}_{f}\}_{*_{A}+}|\psi_{oc}>=\delta^{mm'}|\psi^{m'}_{f}>$, which means that
$(\hat{b}^{m }_{f}{}_{*_A} \hat{b}^{m' \dagger}_{f}+
\hat{b}^{m' \dagger}_{f}{}_{*_A} \hat{b}^{m }_{f})_{*_{A}}|\psi_{oc}>=
\delta^{mm'}|\psi^{m'}_{f}>$. This proofs follows from a. and b..\\

Allowing however, that also $f\ne f'$, one finds that
% $(\hat{b}^{m }_{f}{}_{*_A} \hat{b}^{m' \dagger}_{f}{}_{*_A} |\psi_{oc}>
%= \delta^{mm'}|\psi^{oc}>$, while
the term $\hat{b}^{m }_{f}{}_{*_A} \hat{b}^{m' \dagger}_{f'}{}_{*_A}
|\psi_{oc}>$ gives a nonzero contribution --- what is an even Clifford object and not
zero. But after reducing the Clifford algebra space,
Eq.~(\ref{tildegammareduced}), the two irreducible representations $(f,f')$
carry two different family quantum numbers. The algebraic product has no
meaning any longer: Two different families, reachable from each other
by $\tilde{S}^{ab}$, are orthogonal in the sense of the tensor product.

% Holger, complain!

(As an example let us demonstrate this on the case $d=(5+1)$, presented on
Table~\ref{cliff basis5+1.}: $\hat{b}^{m =1}_{f=II}=
{\scriptstyle (-)} \stackrel{56}{(-)}|\stackrel{12}{[+]} \stackrel{03}{[+i]}$,
$\hat{b}^{m=1 \dagger }_{f=I}= \stackrel{03}{(+i)}\,\stackrel{12}{(+)}| \stackrel{56}{(+)}$. One obtains $\hat{b}^{m =1}_{f=II}{}_{*_A}
\hat{b}^{m =1 \dagger}_{f=I}{}_{*_A} \,|\psi_{oc}>= \stackrel{03}{(+i)}\,\stackrel{12}{(+)}\stackrel{56}{[-]}$, what represents  a even Clifford object and not
%from the algebraic point of view
a nonzero contribution.)

%* change notation on Table.Note that the family quantum number changes
%  situation!

%*03.01.2021 at 19:30

%*TU Check to what part this belongs!

(c.i. Let us show as an example, that $\hat{b}^m_{f}{}_{*_A}\hat{b}^{m'}_{f}=0$.
Each annihilation operator of the same irreducible representation $f$ and
different member $m'$ follows from $\hat{b}^m_{f}{}$,
$\hat{b}^{m'}_{f}=(S^{m' m} \hat{b}^{m\dagger}_{f})^{\dagger} =$
$\hat{b}^{m}_{f} (S^{m m'})^{\dagger}$. Correspondingly we have
\hat{b}^m_{f}{}_{*_A}\hat{b}^{m}_{f} (S^{m m'})^{\dagger}=0$.)\\

{\bf Statement 8:} $
<\bf {\psi^{s' }}_{f'} (\vec{p'})| \bf {\psi^{s }}_{f} (\vec{p})>=$
$\delta^{s s'} \delta_{f f'}\,\delta(\vec{p}'-\vec{p})$.

Let us prove this relation.

Since the ``basis vectors'' in internal space of fermions are orthogonal according to
\begin{eqnarray}
\label{ccorthogonal0app}
\{\hat{b}^{ m}_{f}\,{}_{*_{A}}\,,\,
\hat{b}^{ m' \dagger}_{f'}\,{}_{*_{A}}\}_{+}|\psi_{oc}>&=&
\hat{b}^{ m}_{f}\;{}_{*_{A}}\,
\hat{b}^{ m' \dagger}_{f'}\,{}_{*_{A}}|\psi_{oc}> = \delta^{m m'} \delta_{f f'}\,
\end{eqnarray}
it follows by taking into account Eq.~(\ref{usefulcontinuous}) of
App.~\ref{continuous}~\footnote{In Eq.~(\ref{usefulcontinuous}) one finds
the relation $<\vec{p}|f^{*}(\hat{\vec{p}})\, f(\hat{\vec{p\,'}})|\vec{p\,'}>=
f^{*}(\hat{\vec{p}})\, f(\hat{\vec{p\,'}}) \delta(\vec{p}-\vec{p\,'})$.},
here we
leave out the tensor product between states of ordinary space and internal space
of fermions.
\begin{eqnarray}
\label{ccorthogonal01app}
&&<0_{\vec{p}}|<\psi_{oc}|\hat{\bf b}^{s'}_{f'} (\vec{p}')
\hat{\bf b}^{s  \dagger}_{f} (\vec{p}) |\psi_{oc}>|0_{\vec{p}}>
\nonumber\\
&&= <0_{\vec{p}}|<\psi_{oc}|\sum_{m,m'} c^{s'm' *}{}_{f'} (\vec{p}') \,
\hat{b}_{\vec{p}'} \,\hat{b}^{m'}_{f'}\, c^{s m}{}_{f} (\vec{p}) \,
\hat{b}^{m \dagger }_{f}\,\hat{b}^{\dagger}_{\vec{p}}
\,|\psi_{oc}>|0_{\vec{p}}>\nonumber\\
&&=\sum_{m,m'} <\psi_{oc}|\hat{b}^{m'}_{f'} {}*_{A}
\hat{b}^{m  \dagger}_{f}  \,|\psi_{oc}>
<0_{\vec{p}}| c^{s'm' *}{}_{f'} (\vec{p}') \, \hat{b}_{\vec{p}'} \,
\hat{b}^{\dagger}_{\vec{p}} \,c^{s m}{}_{f} (\vec{p}) |0_{\vec{p}}>=\nonumber\\
&&\sum_m\,  c^{ms *}{}_{f}  (\vec{p},|p^0|=|\vec{p}| )\;\;
c^{m s' }{}_{f'} (\vec{p},|p^0|=|\vec{p}| )\,
\delta_{f f'} \,\delta(\vec{p}'-\vec{p})\nonumber\\
&&= \delta^{s s'} \delta_{f f'}\,\delta(\vec{p}'-\vec{p})\,,
\end{eqnarray}
since $<\psi_{oc}|\hat{b}^{m'}_{f'} {}*_{A}
\hat{b}^{m  \dagger}_{f}  \,|\psi_{oc}>=\delta^{m m'} \delta_{f f'}$
%since the ''basis vectors'' in internal space of fermions are orthogonal according to
%Eq.~(\ref{almostDirac}), \\
%$\{\hat{b}^{ m}_{f}\,{}_{*_{A}}\,,\,
%\hat{b}^{ m' \dagger}_{f'}\,{}_{*_{A}}\}_{+}|\psi_{oc}>=$
%$\hat{b}^{ m}_{f}\;{}_{*_{A}}\, \hat{b}^{ m' \dagger}_{f'}\,{}_{*_{A}}|\psi_{oc}>$
%  $= \delta^{m m'} \delta_{f f'}\,|\psi_{oc}>$,
$<0_{\vec{p}}|\hat{b}_{\vec{p}'} \hat{b}^{\dagger}_{\vec{p}} |0_{\vec{p}}>=
\delta(\vec{p}'-\vec{p})$,  App.~\ref{continuous},  while
%$ {\rm it \; follows \;for\; particular\;}  \vec{p}\,, p^0=
%|\vec{p}|,{\;\rm that} $
\begin{eqnarray}
\label{ccorthogonal1app}
&& \sum_{m} c^{ms *}{}_{f}  (\vec{p},|p^0|=|\vec{p}| )\;\;
c^{m s' }{}_{f'} (\vec{p},|p^0|=|\vec{p}| ) =
\delta^{s s'} \delta_{f f'}\, \,.
\end{eqnarray}
%The orthogonality relation of Eq.~(\ref{ccorthogonal1}) would be comparable
%with the  orthogonality relation of Eq.~(\ref{vvdagger}), provided that we take
%into account the orthogonality of momentum basis, discussed in

%*03.02.2021 at 21:00

\section{Continuous spectra in momentum and coordinate space}
\label{continuous}

%Corrected 09.03.2021

% {\bf Drop comments after taking them into account!}\\

% \noindent
% {\bf HN: }It might help to be definite about the meaning of the symbols,
% e.g.I think you have $\hat{b}_{\vec{p}}=\exp(i\vec{p}\cdot \vec{x})$
% in $\vec{x}$-representation, so that operating in the single particle space
% by $\hat{b}_{\vec{p}}$ one shifts the momentum by the amount
% $\vec{p}$.
% I am now trying to guess what the meaning of $|{\bf O}_{\vec{p}}>$ is:
% I think it is the single particle state of only the posion-momentum
% degrees of freedom with $\vec{p} =\vec{0}$. In position representation it
% means a constant.\\
% {\bf HN:}You seemingly pretend that you can normalize yourself out of
% the usual $2\pi$ comming in. I would in the left equation in (7) have put
% in $(2\pi)^{d-1}$ in front of the delta functions, and then compensatingly
% put $(2\pi)^{d-1}$
%  in the denominator in (8). I doubt you can normalize so
% as to avoid that.

Creation  and annihilation operators for a single fermion state must include besides
the internal space of fermions, described by the superposition of odd products
of Clifford algebra operators $\gamma^a$'s, denoted by
$\hat{b}^{m \dagger}_{f}$, also  the  momentum or coordinate part. We make
the tensor product, $\,*_{T}\,$, of both parts, of the anticommuting internal one
and the commuting momentum or coordinate one, as suggested in
\begin{eqnarray}
\label{wholespacegeneralapp}
\{{\bf \hat{b}}^{m \dagger}_{f} (\vec{p}) \,&=&
\hat{b}^{m \dagger}_{f}\,\,*_{T}\, \hat{b}^{\dagger}_{\vec{p}}\}
% c^{s m}{}_{f} (\vec{p})\}
\; |\psi_{oc}>\,*_{T}\, |0_{\vec{p}}> \,.
\end{eqnarray}
The quantum numbers $(m,f)$ represent  the internal part of anticommuting
"basis states" $\hat{b}^{m \dagger}_{f}$ of the family member $m$ of the
family $f$,
the vacuum state written here as $|\psi_{oc}>\,*_{T}\, |0_{\vec{p}}>$ is
the vacuum for the starting single particle state, from which one obtains
other single particle states with the same internal part, by the operators,
$\hat{b}^{\dagger}_{\vec{p}}$, which pushes the momentum by an amount
%we have to study the continuous spectra in the ordinary space.
%  with the state written here as $|\psi_{oc} >\,*_{T} |0_{\vec{p}} >$ being
% considered as the vacuum for the starting single particle state from which
% you obtain the other single particle states with the same internal part,
% denoted with two quantum numbers, $(m,f)$,  by the operators, like
% $\hat{b}_{\vec{p}}$, which pushes the momentum by an
%   amount $\vec{p}$.

We study  free massless fermions in ($d=(d-1)+1$)-dimensional space,
for any  $d$.
We treat  the continuous spectrum of  $\vec{p}$ assuming
that solutions of the Euler-Lagrange equations relate momentum $\vec{p}$
and energy $p^0$, $(p^0)^2 =(\vec{p})^2$.

We follow up to Eq.~(\ref{eigenvalue}) more or less Ref.~[146], while
neither  Eq.~(\ref{eigenvalue}) nor the rest of equations are from
Ref.~[146] or we have found in  the literature.

The commutation relations of the Hermitian operators $\hat{\vec{x}}$
and $\hat{\vec{p}}$, $\hat{\vec{x}}^{\dagger}=\hat{\vec{x}}$
and  $\hat{\vec{p}}^{\dagger}=\hat{\vec{p}}$,
in units in which $\hbar=1=c$,   are as follows
\begin{eqnarray}
\label{compx0}
\{\hat{x}^i\,, \hat{x}^j \}_- &=& 0\,, \quad
\{\hat{x}^i\,, \hat{p}^j \}_-   =  \delta^{ij}\,,\quad
\{\hat{p}^i\,, \hat{p}^j \}_-   =  0\,.\quad
\end{eqnarray}
Let us  write down some useful relations, following from Eq.~(\ref{compx0}),
\begin{eqnarray}
\label{compx1}
\{\hat{\vec{x}}\,, f(\hat{\vec{x}}, \hat{\vec{p}})\}_- &=&
i \frac{\partial f(\hat{\vec{x}}, \hat{\vec{p}})}{\partial \hat{\vec{p}}}\,, \quad
\{\hat{\vec{p}}\,, f(\hat{\vec{x}}, \hat{\vec{p}})\}_-   =
- i \frac{\partial f(\hat{\vec{x}}, \hat{\vec{p}})}{\partial \hat{\vec{x}}}\,,
\end{eqnarray}
Let us solve the eigenvalue equations for a state $| \vec{p}>$  and $| \vec{x}>$,
both changing continuously with  $\vec{p}$ and $\vec{x}$, respectively,
\begin{eqnarray}
\label{eigenvalue0}
\hat{\vec{p}}\,| \vec{p}>&=& \vec{p}\,| \vec{p}>\,,\qquad \,
\hat{\vec{x}}\,| \vec{x}>  = \vec{x}\,| \vec{x}>\,,\nonumber\\
| \vec{p}>&=& e^{i \hat{\vec{x}}\cdot \vec{p}}\, |0_{\vec{p}}>\,,\quad\;
| \vec{x}>  =  e^{-i \hat{\vec{p}}\cdot \vec{x}}\, |0_{\vec{x}}>\,,\nonumber\\
< \vec{p}|&=& <0_{\vec{p}}|\,e^{-i \hat{\vec{x}}\cdot \vec{p}}\, ,\quad\;
< \vec{x}|  =  <0_{\vec{x}}|\,e^{i \hat{\vec{p}}\cdot \vec{x}}\, ,
\end{eqnarray}
the operator $ e^{i \hat{\vec{x}}\cdot \vec{p}}\, $ translates $|0_{\vec{p}}>$
to $|\vec{p}>$ and operator $e^{-i \hat{\vec{p}}\cdot \vec{x}}$ translates
$|0_{\vec{x}}>$ to $|\vec{x}>$, where $|0_{\vec{p}}>$ and $|0_{\vec{x}}>$
are states with eigenvalues  $\vec{0}$ in both cases.

The continuous spectra are normalized as follows
\begin{eqnarray}
\label{eigenvalueorth}
<\vec{p}\,|\, \vec{p'}>&=& \delta(\vec{p}- \vec{p'})=
\prod_{i=1, d-1} \delta(p^{i}-p^{'i})\,,\quad
<\vec{x}\,|\, \vec{x'}>  = \delta(\vec{x}- \vec{x'})=
\prod_{i=1, d-1} \delta(x^{i}-x^{'i})\,,
\end{eqnarray}
and fulfil the completeness relations
\begin{eqnarray}
\label{eigenvalueorthcom}
\int |\vec{p}> d^{d-1} p <\vec{p}| &=& I\,, \quad
\int |\vec{x}> d^{d-1} x <\vec{x}|   = I\,.\\
\end{eqnarray}

Due to the fact that $\hat{\vec{p}}$ and $\hat{\vec{x}}$ do not commute, Eq.~(\ref{compx0}), we have to evaluate $<\vec{p}| \vec{x}>$, using Eq.~(\ref{eigenvalue0}) and the relations $ f(\hat{\vec{p}}) \,| \vec{p}>=
f(\vec{p}) \,| \vec{p}>$ and $ f(\hat{\vec{x}}) \,| \vec{x}>=
f(\vec{x}) \,| \vec{x}>$. It follows
\begin{eqnarray}
\label{px0}
<\vec{p}\,|\, \vec{x}>&=& <\vec{p}\,| e^{-i \hat{\vec{p}}\cdot \vec{x}}
|0_{\vec{x}}> = e^{-i \vec{p}\cdot \vec{x}} <\vec{p}\,|\, 0_{\vec{x}}>
\nonumber\\
e^{-i \vec{p}\cdot \vec{x}} <0_{\vec{p}}\,|\,
e^{-i \hat{\vec{x}}\cdot \vec{p}}|0_{\vec{x}}>=e^{-i \vec{p}\cdot \vec{x}}
<0_{\vec{p}}\,|\, 0_{\vec{x}}>\,.
\end{eqnarray}
Taking into account $<\vec{p}\,|\, \vec{x}>^*=<\vec{x}\,|\, \vec{p}>$, and
the orthogonality and completeness relations, we
evaluate the relation
$|<0_{\vec{p}}\,|\, 0_{\vec{x}}>|=\frac{1}{(\sqrt{2\pi})^{d-1}}$,
\begin{eqnarray}
\label{px1}
\int <\vec{p}\,| \vec{x}> d\vec{x} <\vec{x}\,| \vec{p'}>&=&
\delta(\vec{p}- \vec{p'}) = |<0_{\vec{p}}|0_{\vec{x}}>|^2
\int  e^{-i (\vec{p}-\vec{p}')\cdot \vec{x}} d\vec{x}\nonumber\\
& =& |<0_{\vec{p}}|0_{\vec{x}}>|^2
\end{eqnarray}
from where we find
\begin{eqnarray}
\label{px2}
<\vec{p}| \vec{x}>&=&e^{i \vec{p}\cdot \vec{x}} \frac{1}{\sqrt{(2 \pi)^{d-1}}}
=<\vec{x}| \vec{p}>^*
\end{eqnarray}

We started this section with the intention to define the creation operators
for fermions, which would include both spaces, the internal one and the
momentum (or coordinate) one, as presented in

Therefore we rewrite Eqs.~(\ref{eigenvalue0}, \ref{eigenvalueorth}) as follows
\begin{eqnarray}
\label{eigenvalue}
| \vec{p}>&=& \hat{b}^{\dagger}_{\vec{p}}\,|0_{\vec{p}}>=
\hat{b}_{-\vec{p}}\,|0_{\vec{p}}>\,,\qquad \,
| \vec{x}>  = \hat{b}^{\dagger}_{\vec{x}}\,| 0_{\vec{x}}>=
\hat{b}_{-\vec{x}}\,| 0_{\vec{x}}>\,,\nonumber\\
< \vec{p}|&=& <0_{\vec{p}}|\,\hat{b}_{\vec{p}}\,=
<0_{\vec{p}}|\, \hat{b}^{\dagger}_{-\vec{p}}\,,\qquad \,
%| \vec{p}>&=& e^{i \hat{\vec{x}}\cdot \vec{p}}\, |0_{\vec{p}}>\,,\quad\;
< \vec{x}|  = <0_{\vec{x}}| \, \hat{b}_{\vec{x}} =
<0_{\vec{x}}| \, \hat{b}^{\dagger}_{-\vec{x}} \,,\nonumber\\
< \vec{p}| \vec{p'}>&=& <0_{\vec{p}}|\,\hat{b}_{\vec{p}}\,
\hat{b}^{\dagger}_{\vec{p'}}\,|0_{\vec{p}}>= \delta(\vec{p}-\vec{p'})\, , \;
< \vec{x}| \vec{x'}>= <0_{\vec{x}}|\,\hat{b}_{\vec{x}}\,
\hat{b}^{\dagger}_{\vec{x'}}\,|0_{\vec{x}}>= \delta(\vec{x}-\vec{x'})\, ,
\nonumber\\
\hat{b}_{\vec{p}}\,\hat{b}^{\dagger}_{\vec{p'}}&=&\delta(\vec{p}-\vec{p'})\,,
\qquad