text:
\section*{Introduction}

In this paper we prove isomorphisms between five Lie groups in complexified Clifford algebra and classical matrix groups in the case of arbitrary dimension and arbitrary signature. Also we obtain isomorphisms of corresponding Lie algebras. We further develop results of the paper \cite{first}.
In \cite{first} you can find statements only in the cases of fixed signatures when the corresponding real Clifford algebra has faithful and irreducible representations over ${\Bbb R}$ and ${\Bbb R}\oplus{\Bbb R}$. In the present paper we prove isomorphisms in all remaining cases, when real Clifford algebra has faithful and irreducible representations over ${\Bbb C}$, ${\Bbb H}$, and ${\Bbb H}\oplus{\Bbb H}$.

Lie groups which are considered in this paper can be useful in different questions of field theory. Note that the following three groups ${\rm G}^{2i1}_{p,q}$, ${\rm G}^{23}_{p,q}$, ${\rm G}^{2}_{p,q}$ are subgroups of pseudo-unitary group (see \cite{pseudou}, \cite{Snygg}, \cite{Dyab}). In \cite{Marchukeng} new symmetry of Dirac equation \cite{Dirac}, \cite{Dirac2} with respect to the pseudo-unitary group was considered.
Spin group ${\rm Spin}_{+}(p,q)$ is a subgroup of all five considered Lie groups. Moreover, group ${\rm Spin}_{+}(p,q)$ coincides with group ${\rm G}^{2}_{p,q}$ in the cases of dimensions $n\leq 5$. We discuss it in details in the Section \ref{SectionSpin}. Salingaros vee group (it consists of $\pm$ basis elements of real Clifford algebra) \cite{Sal1}, \cite{Sal2}, \cite{Sal3}, \cite{Helm2}, \cite{Abl2}, \cite{Abl3} is a subgroup of spin group and, so, also of five considered groups.

We mention a series of articles \cite{Abl1}, \cite{Abl2}, \cite{Abl3}, where many interesting facts about so-called transposition anti-involution are discussed. We consider such conjugation in real and complexified Clifford algebras and call it Hermitian conjugation in \cite{Marchuk:Shirokov}. In \cite{Marchuk:Shirokov} we were interested, mostly, in some particular problems related to applications in field theory; in \cite{Abl1}, \cite{Abl2}, \cite{Abl3} you can find more detailed description of corresponding algebraic structures. Note that operation of Hermitian conjugation of Clifford algebra elements is well-known, especially, in particular cases. For example, P. Dirac \cite{Dirac}, \cite{Dirac2} uses it in the case of signature $(p,q)=(1,3)$ in the theory of Dirac equation for electron. We should note that information about Hermitian conjugation in the Section 2 of the present paper is related to results of the paper \cite{Abl2}. In particular, one finds information about connection between Hermitian conjugation (or so-called transposition anti-involution) and matrix operations of corresponding matrix representations: in \cite{Abl2} for representations based on the fixed idempotent and the basis of corresponding left ideal, in the present paper (see formulas (\ref{relat})-(\ref{3rt})) for fixed matrix representations, in \cite{Marchuk:Shirokov} for complexified Clifford algebras and their representations based on the fixed idempotent and the basis of corresponding left ideal. In \cite{Abl2}, \cite{Abl3} some interesting facts about group $G_{p,q}^\epsilon=\{U\in{C}\!\ell_{p,q}\, | \, U^\dagger U=e\}$ in real Clifford algebras were considered. In \cite{Marchuk:Shirokov} we consider analogue of this group $\{U\in{\Bbb C}\otimes{C}\!\ell_{p,q}\, | \, U^\dagger U=e\}$ (so-called unitary group) in complexified Clifford algebras. The group $G_{p,q}^\epsilon$ is subgroup of this group. Note, that group $G^\epsilon_{p,q}$ coincides in particular cases with some of groups which are considered in the present paper: with the group ${\rm G}^{23}_{p,q}$ in the case of signature $(n,0$), with the group ${\rm G}^{12}_{p,q}$ in the case of signature $(0,n)$. Note, that some of groups which are considered in the present paper are related to automorphism groups of the scalar products on the spinor spaces (see \cite{Port}, \cite{Lounesto}, \cite{Abl3}), but we do not use this fact in the present paper. In \cite{Lounesto} one finds isomorphisms between groups ${\rm G}^{12}_{p,q}$, ${\rm G}^{23}_{p,q}$ and classical matrix Lie groups. In the present paper we also obtain these isomorphisms and also isomorphisms for groups ${\rm G}^{2i1}_{p,q}$, ${\rm G}^{2i3}_{p,q}$, ${\rm G}^{2}_{p,q}$ using other techniques based on relations between operations of conjugations in Clifford algebras and corresponding matrix operations. In particular, we generalized the notion of additional signature \cite{Pauli2} to the case of real Clifford algebras. We also study the corresponding Lie algebras with the use of techniques of quaternion types \cite{QuatAaca}, \cite{Quat2Aaca}, \cite{DAN} in Clifford algebras.

Let us consider the real Clifford algebra ${C}\!\ell_{p,q}$ and complexified Clifford algebra ${\Bbb C}\otimes{C}\!\ell_{p,q}$, $p+q=n$, $n\geq1$ \cite{Clifford}.
The construction of real and complexified Clifford algebras is discussed in details in \cite{Lounesto}, \cite{Chevalley}, \cite{Helm}.

Let us remind the basic notation. Let $e$ be the identity element and let $e^a$, $a=1,\ldots,n$ be generators\footnote{Note that $e^a$ is not exponent. We use notation with upper indices \cite{Benn:Tucker}.} of the Clifford algebra ${C}\!\ell_{p,q}$,
$e^a e^b+e^b e^a=2\eta^{ab}e$, where $\eta=||\eta^{ab}||$ is the diagonal matrix with $+1$ appearing $p$ times on the diagonal and $-1$ appearing $q$ times on the diagonal. Elements
$e^{a_1\ldots a_k}=e^{a_1}\cdots e^{a_k}$, $a_1<\cdots<a_k$, $k=1,\ldots,n$, together with the identity element $e$, form a basis of Clifford algebra. Any Clifford algebra element $U\in{C}\!\ell_{p,q}$ can be written in the form\footnote{We use Einstein's summation convention: there is a sum over index $a$.}
\begin{eqnarray}
U=ue+u_a e^a+\sum_{a_1<a_2}u_{a_1 a_2}e^{a_1 a_2}+\cdots+u_{1\ldots n}e^{1\ldots n},\label{uform}
\end{eqnarray}
where $u, u_a, u_{a_1 a_2}, \ldots, u_{1\ldots n}$ are real numbers. For arbitrary element $U\in{\Bbb C}\otimes{C}\!\ell_{p,q}$ of complexified Clifford algebra we use the same notation (\ref{uform}), where $u, u_a, u_{a_1 a_2}, \ldots, u_{1\ldots n}$ are complex numbers.

 We denote by ${C}\!\ell^k_{p,q}$ the vector spaces that span over the basis elements
$e^{a_1\ldots a_k}$. Elements of ${C}\!\ell^k_{p,q}$ are said to be
elements of grade $k$. We have ${C}\!\ell_{p,q}=\bigoplus_{k=0}^{n}{C}\!\ell^k_{p,q}.$
Clifford algebra is a $Z_2$-graded algebra and it is represented as the direct sum of even and odd subspaces:
$${C}\!\ell_{p,q}={C}\!\ell^{(0)}_{p,q}\oplus{C}\!\ell^{(1)}_{p,q},\quad {C}\!\ell^{(i)}_{p,q}{C}\!\ell^{(j)}_{p,q}\subseteq{C}\!\ell^{(i+j)\,\rm{mod} 2}_{p,q},\quad \mbox{where}\quad {C}\!\ell^{(i)}_{p,q}=\bigoplus_{k\equiv i\,\rm{mod}2}{C}\!\ell^k_{p,q},\quad i, j=0, 1.$$

Let us consider the Clifford algebra ${C}\!\ell_{p,q}$ as the vector space and represent it in the form of the direct sum of four subspaces of {\it quaternion types} 0, 1, 2 and 3 (see \cite{QuatAaca}, \cite{Quat2Aaca}, \cite{DAN}):
$${C}\!\ell_{p,q}=\overline{\textbf{0}}\oplus\overline{\textbf{1}}\oplus\overline{\textbf{2}}\oplus\overline{\textbf{3}},\qquad
\mbox{where}\quad \overline{\textbf{s}}=\bigoplus_{k\equiv s\,\rm{mod} 4}{C}\!\ell^k_{p,q},\quad s=0, 1, 2, 3.$$
 We represent complexified Clifford algebra ${\Bbb C}\otimes{C}\!\ell_{p,q}$ in the form of the direct sum of eight subspaces:
${\Bbb C}\otimes{C}\!\ell_{p,q}=\overline{\textbf{0}}\oplus\overline{\textbf{1}}\oplus\overline{\textbf{2}}\oplus\overline{\textbf{3}} \oplus i\overline{\textbf{0}}\oplus i\overline{\textbf{1}}\oplus i\overline{\textbf{2}}\oplus i\overline{\textbf{3}}.$

In \cite{first} we discussed a recurrent method of construction of matrix representation of real Clifford algebra in the cases of signatures $p-q\equiv0, 1, 2\mod 8$. In the Section \ref{sectmatr} of the present paper we generalize this method to the case of arbitrary signature.

In the Section \ref{sectherm} of the present paper we give some information about Hermitian conjugation in real and complexified Clifford algebras (see also \cite{Marchuk:Shirokov} and \cite{Abl2} for the case of real Clifford algebras). In \cite{Pauli2} we introduced the notion of additional signature $(k,l)$ of complexified Clifford algebra. In the Section \ref{sectadd} of the present paper  we generalize this notion to the case of real Clifford algebras.

In the Section \ref{sectmain} we prove isomorphisms between five Lie groups in Clifford algebra and classical matrix Lie groups (some of these isomorphisms are known, see above). We study corresponding Lie algebras. In the Section \ref{SectionSpin} of the present paper we discuss connection between groups ${\rm Spin}_{+}(p,q)$ and ${\rm G}^{2}_{p,q}$. In the Section \ref{sectconcl} we summarize results of \cite{first} and the present paper.

\section{Recurrent method of construction of matrix representations of real Clifford algebras in the case of arbitrary signature}\label{sectmatr}

In \cite{first} we discussed a recurrent method of construction of matrix representation only in the cases of signatures $p-q\equiv0, 1, 2\mod 8$ (the case of faithful and irreducible representations over ${\Bbb R}$ and ${\Bbb R}\oplus{\Bbb R}$). Now we generalize this method to the case of arbitrary signature. Clifford algebra has faithful and irreducible representations over ${\Bbb R}$, ${\Bbb R}\oplus{\Bbb R}$, ${\Bbb C}$, ${\Bbb H}$, ${\Bbb H}\oplus{\Bbb H}$ in different cases. However, items 1-4 are the same as in \cite{first}.

We have the following well-known isomorphisms ${C}\!\ell_{p,q}\simeq {\rm L}_{p,q}$, where we denote by ${\rm L}_{p,q}$ the following matrix algebras
\begin{equation}
{\rm L}_{p,q}=\left\lbrace
\begin{array}{ll}
{\rm Mat}(2^{\frac{n}{2}},{\Bbb R}), & \parbox{.5\linewidth}{ if $p-q\equiv0; 2\!\!\mod 8$;}\\
{\rm Mat}(2^{\frac{n-1}{2}},{\Bbb R})\oplus {\rm Mat}(2^{\frac{n-1}{2}},{\Bbb R}), & \parbox{.5\linewidth}{ if $p-q\equiv1\!\!\mod 8$;}\\
{\rm Mat}(2^{\frac{n-1}{2}},{\Bbb C}), & \parbox{.5\linewidth}{ if $p-q\equiv3; 7\!\!\mod 8$;}\\
{\rm Mat}(2^{\frac{n-2}{2}},\mathbb H), & \parbox{.5\linewidth}{ if $p-q\equiv4; 6\!\!\mod 8$;}\\
{\rm Mat}(2^{\frac{n-3}{2}},\mathbb H)\oplus {\rm Mat}(2^{\frac{n-3}{2}},\mathbb H), & \parbox{.5\linewidth}{ if $p-q\equiv5\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}

We want to construct faithful and irreducible matrix representation $\beta: {C}\!\ell_{p,q}\to {\rm L}_{p,q}$ of all real Clifford algebras with some additional properties related to symmetry and skew-symmetry of corresponding matrices (we discuss it in the next section). In some particular cases we construct $\beta$ in the following way:
\begin{itemize}
  \item In the case ${C}\!\ell_{0,0}$: $e\to 1$.
  \item In the case ${C}\!\ell_{0,1}$: $e\to 1$, $e^1\to i$.
  \item In the case ${C}\!\ell_{1,0}$: $e\to {\bf1}={\rm diag}(1,1)$, $e^1\to {\rm diag}(1,-1)$.
  \item In the case ${C}\!\ell_{0,2}$: $e \to 1$, $e^1 \to i$, $e^2 \to j$.
  \item In the case ${C}\!\ell_{0,3}$: $e\to {\bf1}={\rm diag}(1,1)$, $e^1 \to {\rm diag}(i, -i)$, $e^2 \to {\rm diag}(j,-j)$, $e^3 \to {\rm diag}(k,-k)$.
\end{itemize}
For basis element $e^{a_1 \ldots a_k}$ we use the matrix that equals the product of matrices corresponding to generators $e^{a_1}, \ldots, e^{a_k}$. For identity element $e$, we always use identity matrix {\bf 1} of corresponding size.

Suppose that we have a faithful and irreducible matrix representation $\beta:{C}\!\ell_{p,q}\to {\rm L}_{p,q}$
\begin{eqnarray}
e^a \to \beta^a,\quad a=1, \ldots, n,\label{matpred}
\end{eqnarray}
where $\beta^a=\beta(e^a)$, $a=1, 2, \ldots, n$. Now we want to construct the matrix representations of other Clifford algebras ${C}\!\ell_{p+1, q+1}$, ${C}\!\ell_{q+1, p-1}$, ${C}\!\ell_{p-4, q+4}$ with generators $e^a$ ($a=1, 2, \ldots, n+2$ in the first case and $a=1, 2, \ldots, n$ in the last two cases) with the use of matrices $\beta^a$.  Using the following items 1-4 and Cartan periodicity of real Clifford algebras we obtain matrix representations of all real Clifford algebras. We call it recurrent method of construction of matrix representations\footnote{Another method of construction of matrix representations is based on the idempotent and basis of corresponding left ideal (see, for example, \cite{Abl1}, \cite{Abl2}, \cite{Abl3}, \cite{Marchuk:Shirokov}).}.

1.\quad Let us consider ${C}\!\ell_{p+1, q+1}$. If $p-q\not\equiv 1\!\!\mod 4$, then for $p$ generators with squares $+1$ and $q$ generators with squares $-1$ we have
$$e^a\to \left( \begin{array}{ll}
 \beta^a & 0 \\
 0 & -\beta^a \end{array}\right),\qquad a=1, 2, \ldots, p, p+2, p+3, \ldots, p+q+1.$$
And for two remaining generators we have
$$
 e^{p+1}\to\left( \begin{array}{ll}
 0 & {\bf 1} \\
 {\bf 1} & 0 \end{array}\right),\quad
 e^{p+q+2}\to\left( \begin{array}{ll}
 0 & -{\bf 1} \\
 {\bf 1} & 0 \end{array}\right).
 $$

2. \quad If $p-q\equiv1\!\!\mod 4$, then matrices (\ref{matpred}) are block-diagonal and we have the following matrix representation of ${C}\!\ell_{p+1, q+1}$. For $p+q$ generators $e^a$, $a=1, 2, \ldots, p, p+2, p+3, \ldots, p+q+1$ we have the same as in the previous item and for remaining two generators we have
 $$e^{p+1}\to
 \left( \begin{array}{ll}
 \beta^1\ldots\beta^n\Omega & 0 \\
 0 & -\beta^1\ldots\beta^n\Omega
  \end{array}\right),\quad
 e^{p+q+2}\to\left( \begin{array}{ll}
 \Omega & 0 \\
 0 & -\Omega
 \end{array}\right),
 $$
where
$$
\Omega =\left( \begin{array}{ll}
 0 & -{\bf 1} \\
 {\bf 1} & 0 \end{array}\right).
$$

3. \quad We construct matrix representation of ${C}\!\ell_{q+1, p-1}$ using
$$e^1\to \beta^1,\qquad e^i\to \beta^i \beta^1,\quad i=2, \ldots, n.$$

4. \quad We construct matrix representation of ${C}\!\ell_{p-4, q+4}$ using
$$e^i\to \beta^i \beta^1 \beta^2 \beta^3 \beta^4,\quad i=1, 2, 3, 4,\qquad e^j\to \beta^j,\quad j=5, \ldots, n.$$

In \cite{first} we gave some examples in the cases $p-q\equiv 0, 1, 2\mod 8$. Now let us give some examples in the cases of other signatures:

\begin{description}
  \item[${C}\!\ell_{1, 2}$]
 $$
  e^1 \to \left( \begin{array}{ll}
 0 & 1 \\
 1 & 0 \end{array}\right),\quad
 e^2 \to \left( \begin{array}{ll}
 i & 0 \\
 0 & -i \end{array}\right),\quad
 e^3 \to \left( \begin{array}{ll}
 0 & -1 \\
 1 & 0 \end{array}\right).
 $$
 \item[${C}\!\ell_{1, 3}$]
 $$
  e^1 \to \left( \begin{array}{ll}
 0 & 1 \\
 1 & 0 \end{array}\right),
 e^2 \to \left( \begin{array}{ll}
 i & 0 \\
 0 & -i \end{array}\right),
 e^3 \to \left( \begin{array}{ll}
 j & 0 \\
 0 & -j \end{array}\right),
 e^4 \to \left( \begin{array}{ll}
 0 & -1 \\
 1 & 0 \end{array}\right).
 $$
 \item[${C}\!\ell_{4, 0}$]
 $$
  e^1 \to \left( \begin{array}{ll}
 0 & 1 \\
 1 & 0 \end{array}\right),
 e^2 \to \left( \begin{array}{ll}
 0 & i \\
 -i & 0 \end{array}\right),
 e^3 \to \left( \begin{array}{ll}
 0 & j \\
 -j & 0 \end{array}\right),
 e^4 \to \left( \begin{array}{ll}
 -1 & 0 \\
 0 & 1 \end{array}\right).
 $$
 \item[${C}\!\ell_{0, 4}$]
 $$
  e^1 \to \left( \begin{array}{ll}
 k & 0 \\
 0 & -k \end{array}\right),
 e^2 \to \left( \begin{array}{ll}
 -j & 0 \\
 0 & -j \end{array}\right),
 e^3 \to \left( \begin{array}{ll}
 i & 0 \\
 0 & i \end{array}\right),
 e^4 \to \left( \begin{array}{ll}
 0 & k \\
 k & 0 \end{array}\right).
 $$
\end{description}

\section{Relation between operations of conjugation in Clifford algebra and matrix operations}\label{sectherm}

Consider the following well-known involutions in real ${C}\!\ell_{p,q}$ and complexified Clifford algebra ${\Bbb C}\otimes{C}\!\ell_{p,q}$:
$$
\hat{U}=U|_{e^a\to-e^a},\quad
\tilde{U}=U|_{e^{a_1\ldots a_r}\to e^{a_r}\ldots e^{a_1}},
$$
where $U$ has the form (\ref{uform}). The operation $U\to \hat{U}$ is called {\it grade involution} and $U\to \tilde{U}$ is called {\it reversion}.
Also we have operation of {\it complex conjugation}
\begin{eqnarray}
\bar U=\bar u e+\bar u_a e^a+\sum_{a_1<a_2}\bar u_{a_1 a_2}e^{a_1
a_2}+\sum_{a_1<a_2<a_3}\bar u_{a_1 a_2 a_3}e^{a_1 a_2
a_3}+\ldots,\nonumber
\end{eqnarray}
where we take complex conjugation of complex numbers $u_{a_1 \ldots a_k}$. Superposition of reversion and complex conjugation is {\em pseudo-Hermitian conjugation of Clifford algebra elements}\footnote{Pseudo-Hermitian conjugation of Clifford algebra elements is related to pseudo-unitary matrix groups as Hermitian conjugation is related to unitary groups, see \cite{Snygg}, \cite{Marchuk:Shirokov}.}
$$
U^\ddagger=\tilde{\bar{U}}.
$$
In the real Clifford algebra ${C}\!\ell_{p,q}\subset {\Bbb C}\otimes{C}\!\ell_{p,q}$ we have $U^\ddagger=\tilde{U}$, because $\bar U=U$.

Let us consider in complexified ${\Bbb C}\otimes{C}\!\ell_{p,q}$ and real ${C}\!\ell_{p,q}\subset{\Bbb C}\otimes{C}\!\ell_{p,q}$ Clifford algebras the linear operation (involution) $\dagger: {\Bbb C}\otimes{C}\!\ell_{p,q}\to{\Bbb C}\otimes{C}\!\ell_{p,q}$ such that $(\lambda e^{a_1 \ldots a_k})^\dagger=\bar{\lambda} (e^{a_1 \ldots a_k})^{-1}$, $\lambda\in{\Bbb C}$. We call this operation {\it Hermitian conjugation of Clifford algebra elements}\footnote{Note that it is not Hermitian conjugation of matrix but it is related to this operation.}. We use notation $e_a=\eta_{ab}e^b=(e^a)^{-1}=(e^a)^\dagger$ and $e_{a_1\ldots a_k}=(e^{a_1 \ldots a_k})^{-1}=(e^{a_1 \ldots a_k})^\dagger$. This operation is well-known and many authors use it, for example, in different questions of field theory in the case of signature $(p,q)=(1,3)$. For more details, see \cite{Marchuk:Shirokov}, and for the case of real Clifford algebras see \cite{Abl1}, \cite{Abl2}, \cite{Abl3} (so-called transposition anti-involution in real Clifford algebras).

Note that we have the following relation between operation of Hermitian conjugation of Clifford algebra elements $\dagger$ and other operations in complexified Clifford algebra ${\Bbb C}\otimes{C}\!\ell_{p,q}$ (see \cite{Marchuk:Shirokov})
\begin{eqnarray}
U^\dagger&=&e_{1\ldots p}U^\ddagger e^{1 \ldots p},\qquad \mbox{if $p$ is odd},\nonumber\\
U^\dagger&=&e_{1\ldots p}\hat{U}^{\ddagger} e^{1 \ldots p},\qquad \mbox{if $p$ is even},\label{sogldager2}\\
U^\dagger&=&e_{p+1\ldots n}U^\ddagger e^{p+1 \ldots n},\qquad \mbox{if $q$ is even},\nonumber\\
U^\dagger&=&e_{p+1\ldots n}\hat{U}^{\ddagger} e^{p+1 \ldots n},\qquad \mbox{if $q$ is odd}.\nonumber
\end{eqnarray}

We have the following well-known isomorphisms:
\begin{eqnarray}
{\Bbb C}\otimes{C}\!\ell_{p,q}&\simeq& {\rm Mat}(2^{\frac{n}{2}}, {\Bbb C}),\qquad \mbox{if $n$ is even},\nonumber\\
{\Bbb C}\otimes{C}\!\ell_{p,q}&\simeq& {\rm Mat}(2^{\frac{n-1}{2}}, {\Bbb C})\oplus {\rm Mat}(2^{\frac{n-1}{2}}, {\Bbb C}),\qquad \mbox{if $n$ is odd}.\nonumber
\end{eqnarray}

Hermitian conjugation of Clifford algebra elements corresponds to Hermitian conjugation of matrix $\beta(U^\dagger)=(\beta(U))^\dagger$ for faithful and irreducible matrix representations over ${\Bbb C}$ and ${\Bbb C}\oplus{\Bbb C}$ of complexified Clifford algebra, based on the fixed idempotent and basis of corresponding left ideal, see \cite{Marchuk:Shirokov}.

It is not difficult to prove that for matrix representation $\beta$ of real Clifford algebra ${C}\!\ell_{p,q}$ from the previous section we have
\begin{eqnarray}
\beta(U^\dagger)&=&(\beta(U))^T,\qquad p-q\equiv0, 1, 2\!\!\mod 8,\nonumber\\
\beta(U^\dagger)&=&(\beta(U)^\dagger,\qquad p-q\equiv3, 7\!\!\mod 8,\label{relat}\\
\beta(U^\dagger)&=&(\beta(U))^*,\qquad p-q\equiv4, 5, 6\!\!\mod 8,\nonumber
\end{eqnarray}
where $T$ is the operation of matrix transposition, $*$ is the operation of conjugate transpose of matrices of quaternions, $U^\dagger$ is the Hermitian conjugate of a Clifford algebra element $U$, $\beta(U)^\dagger$ is the Hermitian conjugate of the corresponding matrix.
Note, that in \cite{Abl2} analogous statement is proved for matrix representations based on the fixed idempotent and the basis of corresponding left ideal. In this paper we do not use idempotent to construct matrix representation. We use the recurrent method of construction of matrix representation (see previous section). So, we must verify statement for fixed matrices in some cases of small $n$ (see the beginning of the previous section) and then, using the method of mathematical induction, we must verify that, using items 1-4, we obtain matrix representations of other Clifford algebras with the same property (\ref{relat}). We omit the detailed proof.

In the cases $p-q\equiv0, 1, 2\!\!\mod 8$ for the real Clifford algebra ${C}\!\ell_{p,q}\subset{\Bbb C}\otimes{C}\!\ell_{p,q}$ we have
\begin{eqnarray}
U^T&=&e_{1 \ldots p} \tilde{U} e^{1 \ldots p},\qquad \mbox{if $p$ is odd},\nonumber\\
U^T&=&e_{1 \ldots p} \tilde{\hat{U}} e^{1 \ldots p},\qquad \mbox{if $p$ is even},\label{0rt}\\
U^T&=&e_{p+1 \ldots n} \tilde{U} e^{p+1 \ldots n},\qquad \mbox{if $q$ is even},\nonumber\\
U^T&=&e_{p+1 \ldots n} \tilde{\hat{U}} e^{p+1 \ldots n},\qquad \mbox{if $q$ is odd,}\nonumber
\end{eqnarray}
where $U^T=\beta^{-1}(\beta^T(U))=U^\dagger$.

In the cases $p-q\equiv3,7\!\!\mod 8$ for the real Clifford algebra ${C}\!\ell_{p,q}$ we have
\begin{eqnarray}
U^\dagger&=&e_{1\ldots p}\tilde{U} e^{1 \ldots p},\qquad \mbox{if $p$ is odd},\nonumber\\
U^\dagger&=&e_{1\ldots p}\tilde{\hat{U}} e^{1 \ldots p},\qquad \mbox{if $p$ is even},\label{1rt}\\
U^\dagger&=&e_{p+1\ldots n}\tilde{U} e^{p+1 \ldots n},\qquad \mbox{if $q$ is even},\nonumber\\
U^\dagger&=&e_{p+1\ldots n}\tilde{\hat{U}} e^{p+1 \ldots n},\qquad \mbox{if $q$ is odd}.\nonumber
\end{eqnarray}

In the cases $p-q\equiv4, 5, 6\!\!\mod 8$ for the real Clifford algebra we have
\begin{eqnarray}
U^*&=&e_{1\ldots p}\tilde{U}e^{1 \ldots p},\qquad \mbox{if $p$ is odd},\nonumber\\
U^*&=&e_{1\ldots p}\tilde{\hat{U}} e^{1 \ldots p},\qquad \mbox{if $p$ is even},\label{3rt}\\
U^*&=&e_{p+1\ldots n}\tilde{U} e^{p+1 \ldots n},\qquad \mbox{if $q$ is even},\nonumber\\
U^*&=&e_{p+1\ldots n}\tilde{\hat{U}} e^{p+1 \ldots n},\qquad \mbox{if $q$ is odd},\nonumber
\end{eqnarray}
where $U^*=\beta^{-1}(\beta^*(U))=U^\dagger$. We use formulas (\ref{0rt}) - (\ref{3rt}) in the next sections.

\section{Additional signature of real Clifford algebra}\label{sectadd}

In \cite{Pauli2} we introduced the notion of additional signature $(k,l)$ of complexified Clifford algebra. Now we want to generalize this notion to the case of real Clifford algebras.

Suppose we have faithful and irreducible matrix representation $\beta$ over ${\Bbb C}$ or ${\Bbb C}\oplus{\Bbb C}$ of complexified Clifford algebra. We can always use such matrix representation that all matrices $\beta^a=\beta(e^a)$ are symmetric or skew-symmetric. Let $k$ be the number of symmetric matrices among $\{ \beta^a \} $ for matrix representation $\beta$, and $l$ be the number of skew-symmetric matrices among $\{ \beta^a \}$. Let $e^{b_1},\ldots, e^{b_k}$ denote the generators for which the matrices are symmetric. Analogously, we have $e^{c_1},\ldots, e^{c_l}$ for skew-symmetric matrices.

We use the notion of additional signature of Clifford algebra when we study relation between matrix representation and operations of conjugation.
In complexified Clifford algebra we have (see \cite{Pauli2})
\begin{eqnarray}
U^T&=&e_{b_1 \ldots b_k} \tilde{U} e^{b_1\ldots b_k},\qquad \mbox{$k$ is odd},\nonumber\\
U^T&=&e_{b_1 \ldots b_k} \tilde{\hat{U}} e^{b_1\ldots b_k},\qquad \mbox{$k$ is even},\label{sogltransp}\\
U^T&=&e_{c_1 \ldots c_l} \tilde{U}e^{c_1\ldots c_l},\qquad \mbox{$l$  is even},\nonumber\\
U^T&=&e_{c_1 \ldots c_l} \tilde{\hat{U}} e^{c_1\ldots c_l},\qquad \mbox{$l$ is odd.}\nonumber
\end{eqnarray}

Numbers $k$ and $l$ depend on the matrix representation $\beta$. But they can take only certain values despite dependence on the matrix representation.

In \cite{Pauli2} we proved that in complexified Clifford algebra we have only the following possible values of additional signature:
$$\begin{array}{ll}
\begin{tabular}{|c|c|}
\hline
$n\mod 8$ & $(k\mod 4, l\mod 4)$   \\ \hline \hline
$0$ & $(0,0),\,(1,3)$  \\ \hline
$1$ & $(1,0)$   \\ \hline
$2$ & $(1,1),\,(2,0)$  \\ \hline
$3$ & $(2,1)$   \\ \hline
\end{tabular} &
\begin{tabular}{|c|c|}
\hline
$n\mod 8$ & $(k\mod 4, l\mod 4)$   \\ \hline \hline
$4$ & $(3,1),\,(2,2)$   \\ \hline
$5$ & $(3,2)$   \\ \hline
$6$ & $(3,3),\,(0,2)$   \\ \hline
$7$ & $(0,3)$   \\ \hline
\end{tabular}
\end{array}
$$

Now we want to use the notion of additional signature in real Clifford algebra. Let us consider faithful and irreducible matrix representation over ${\Bbb R}$, ${\Bbb R}\oplus{\Bbb R}$, ${\Bbb C}$, ${\Bbb H}$, or ${\Bbb H}\oplus{\Bbb H}$ of real Clifford algebra. We can always use such matrix representation that all matrices $\beta^a=\beta(e^a)$ are symmetric or skew-symmetric (for example, we can use matrix representation $\beta:{C}\!\ell_{p,q}\to {\rm L}_{p,q}$ from the Section \ref{sectmatr}). Let $k$ be the number of symmetric matrices among $\{ \beta^a \} $ for matrix representation $\beta$, and $l$ be the number of skew-symmetric matrices among $\{ \beta^a \}$. Let $e^{b_1},\ldots, e^{b_k}$ denote the generators for which the matrices are symmetric. Analogously, we have $e^{c_1},\ldots, e^{c_l}$ for skew-symmetric matrices.

Note, that in the cases $p-q\equiv0, 1, 2\!\!\mod 8$ for the real Clifford algebra ${C}\!\ell_{p,q}$ the formulas (\ref{sogltransp}) are valid and they coincide with formulas (\ref{0rt}) (we have $p=k$ and $q=l$). In the cases $p-q\equiv3,7\!\!\mod 8$ for the real Clifford algebra we can use formulas (\ref{sogltransp}). The proof is similar to the proof for the complexified Clifford algebra. Actually, it is sufficient to prove these formulas for basis elements because of linearity of operation of conjugation. For example, if $k$ is odd, then
\begin{eqnarray}
(e^{a_1 \ldots a_m})^T&=&(e^{b_k})^{-1}\ldots (e^{b_1})^{-1} \tilde{e^{a_1 \ldots a_m}} e^{b_1}\ldots e^{b_k}=\nonumber\\
&=&(e^{b_k})^{-1}\ldots (e^{b_1})^{-1} e^{a_m}\ldots e^{a_1} e^{b_1}\ldots e^{b_k}=\nonumber\\
&=&(-1)^{mk-j} e^{a_m}\ldots e^{a_1}=(-1)^{m-j} e^{a_m}\ldots e^{a_1}\nonumber
\end{eqnarray}
where $j$ is the number of elements $e^{a_1}, \ldots, e^{a_m}$ that have symmetric matrix representation. We can analogously prove remaining three formulas from (\ref{sogltransp}).

But formulas (\ref{sogltransp}) are not valid in real Clifford algebra of signatures $p-q\equiv4, 5, 6\!\!\mod 8$ because of the properties of operation of transposition for quaternionic matrices: $(AB)^T\neq B^T A^T$ for quaternionic matrices $A$ and $B$.

So, it makes sense to use the notion of additional signature of real Clifford algebra only in the cases $p-q\equiv3,7\!\!\mod 8$. We don't want to obtain all possible values of additional signature, but we want to obtain some possible values. If we consider matrix representation $\beta$ from the Section \ref{sectmatr},  then in the case $(p,q)=(0,1)$ we have $(k,l)=(1,0)$. Note, that we use only transformations of types 1, 3 and 4 from Section \ref{sectmatr} to obtain matrix representations of other Clifford algebras of signatures $p-q\equiv3,7\!\!\mod 8$. If we use transformation of type 1, then
$(p,q)\to(p+1, q+1)$ and $(k,l)\to(k+1, l+1)$. When we use transformation of type 3, we have $(p,q)\to(q+1, p-1)$ and $(k,l)\to(l+1, k-1)$ because $\beta^1$ is a symmetric matrix. When we use transformation of type 4, we have $(p,q)\to(p-4,q+4)$ and it is not difficult to obtain that $k$ and $l$ do not change the parity. So, for matrix representation $\beta$ we always have: if $p$ is even and $q$ is odd, then $k$ is odd and $l$ is even (note, that the same is true in complexified Clifford algebra, see table above). For example, for $(p,q)=(1,2)$ we have $(k,l)=(2,1)$, for $(p,q)=(3,0)$ we have $(k,l)=(2,1)$, for $(p,q)=(2,3)$ we have $(k,l)=(3,2)$ and so on. We will use this fact in the proof of Theorem 6.2 in the following section.

\section{Theorems}\label{sectmain}

Let us consider the following subsets of complexified Clifford algebra \cite{first}
\begin{eqnarray}
{\rm G}^{2i1}_{p,q} &=& \{U\in{C}\!\ell^{(0)}_{p,q}\oplus i{C}\!\ell^{(1)}_{p,q} : U^\ddagger U=e\},\nonumber\\
{\rm G}^{2i3}_{p,q} &=& \{U\in{C}\!\ell^{(0)}_{p,q}\oplus i{C}\!\ell^{(1)}_{p,q} : \hat{U}^{\ddagger} U=e\},\nonumber\\
{\rm G}^{23}_{p,q}&=&\{U\in{C}\!\ell_{p,q}\, | \, \tilde{U}U=e\},\label{grou}\\
{\rm G}^{12}_{p,q}&=&\{U\in{C}\!\ell_{p,q}\, | \, \tilde{\hat{U}}U=e\},\nonumber\\
{\rm G}^{2}_{p,q}&=&\{U\in{C}\!\ell^{(0)}_{p,q}\, | \, \tilde{U}U=e\}.\nonumber
\end{eqnarray}
They can be considered as Lie groups. Their Lie algebras are
\begin{eqnarray}
\overline{\textbf{2}}\oplus i\overline{\textbf{1}},\quad \overline{\textbf{2}}\oplus i\overline{\textbf{3}},\quad \overline{\textbf{2}}\oplus\overline{\textbf{3}},\quad \overline{\textbf{2}}\oplus\overline{\textbf{1}},\quad \overline{\textbf{2}}\label{algebr}
\end{eqnarray}
respectively.

Note, that we have the following properties (see \cite{QuatAaca}, \cite{Quat2Aaca}, \cite{DAN})
\begin{eqnarray}
&&[\overline{\textbf{k}},\overline{\textbf{k}}]\subseteq\overline{\textbf{2}},\qquad k=0, 1, 2, 3 \nonumber;\\
&&[\overline{\textbf{k}},\overline{\textbf{2}}]\subseteq\overline{\textbf{k}}, \qquad k=0, 1, 2, 3 \label{1}; \\
&&[\overline{\textbf{0}},\overline{\textbf{1}}]\subseteq\overline{\textbf{3}}, \quad  [\overline{\textbf{0}},\overline{\textbf{3}}]\subseteq\overline{\textbf{1}}, \quad [\overline{\textbf{1}},\overline{\textbf{3}}]\subseteq\overline{\textbf{0}} \nonumber,
\end{eqnarray}
where $[U,V]=UV-VU$ is the commutator of arbitrary
Clifford algebra elements $U$ and $V$.

It is not difficult to calculate dimensions of these Lie algebras, because we know that $\dim {C}\!\ell^k_{p,q}=C_n^k=\frac{n!}{k! (n-k)!}$. So, for example, $$\dim \overline{\textbf{2}}=\sum_{k\equiv 2\!\!\mod 4}C_n^k=2^{n-2}-2^{\frac{n-2}{2}}\cos{\frac{\pi n}{4}}.$$
Let us represent Lie groups, corresponding Lie algebras, and their dimensions in the following table.

\begin{center}
    \begin{tabular}{|c|c|c|}
\hline
Lie group & Lie algebra & dimension   \\ \hline \hline
${\rm G}^{2i1}_{p,q}$ & $\overline{\textbf{2}}\oplus i\overline{\textbf{1}}$ & $2^{n-1}-2^{\frac{n-1}{2}}\cos{\frac{\pi (n+1)}{4}}$  \\ \hline
${\rm G}^{2i3}_{p,q}$ & $\overline{\textbf{2}}\oplus i\overline{\textbf{3}}$ & $2^{n-1}-2^{\frac{n-1}{2}}\sin{\frac{\pi (n+1)}{4}}$   \\ \hline
${\rm G}^{23}_{p,q}$ & $\overline{\textbf{2}}\oplus\overline{\textbf{3}}$ & $2^{n-1}-2^{\frac{n-1}{2}}\sin{\frac{\pi (n+1)}{4}}$  \\ \hline
${\rm G}^{12}_{p,q}$ & $\overline{\textbf{2}}\oplus\overline{\textbf{1}}$ & $2^{n-1}-2^{\frac{n-1}{2}}\cos{\frac{\pi (n+1)}{4}}$  \\ \hline
${\rm G}^{2}_{p,q}$ & $\overline{\textbf{2}}$ & $2^{n-2}-2^{\frac{n-2}{2}}\cos{\frac{\pi n}{4}}$  \\ \hline
\end{tabular}
\end{center}

\begin{thm}\label{theoremggg}\cite{first} We have the following Lie group isomorphisms
$${\rm G}^{2i1}_{p,q}\simeq {\rm G}^{12}_{q,p},\qquad {\rm G}^{2i3}_{p,q}\simeq {\rm G}^{23}_{q,p},\qquad {\rm G}^{2}_{p,q}\simeq {\rm G}^{12}_{p,q-1}\simeq {\rm G}^{12}_{q,p-1},\qquad {\rm G}^{2}_{p,q}\simeq {\rm G}^{2}_{q,p}.$$
\end{thm}

\proof We must use transformation $e^a \to e^a e^n$ or $e^a \to ie^a$, $a=1, 2, \ldots, n$ in different cases. $\blacksquare$

According to Theorem \ref{theoremggg} it is sufficient to consider only groups ${\rm G}^{12}_{p,q}$ and ${\rm G}^{23}_{p,q}$.
In \cite{first} we proved isomorphisms between these groups and classical matrix Lie groups in the cases of signatures $p-q\equiv0, 1, 2\!\!\mod 8$. Now let us consider cases $p-q\equiv3, 7\!\!\mod 8$ and $p-q\equiv4, 5, 6\!\!\mod 8$.

\begin{thm} We have the following Lie group isomorphisms.
In the cases of signatures $p-q\equiv 3, 7\!\!\mod 8$ ($n$ is odd)
\begin{equation}
{\rm G}^{23}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm U}(2^{\frac{n-1}{2}}), & \mbox{$(p,q)=(n,0)$;}\\
{\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), & \parbox{.5\linewidth}{$n\equiv 3, 7\!\!\mod 8$, $q\neq 0$;}\\
{\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 5\!\!\mod 8$;}\\
{\rm O}(2^{\frac{n-1}{2}}, {\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 1\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
In the cases of signatures $p-q\equiv 3, 7\!\!\mod 8$ ($n$ is odd)
\begin{equation}
{\rm G}^{12}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm U}(2^{\frac{n-1}{2}}), & \mbox{$(p,q)=(0,n)$;}\\
{\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), & \parbox{.5\linewidth}{$n\equiv 1, 5\!\!\mod 8$, $p\neq 0$;}\\
{\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 3\!\!\mod 8$;}\\
{\rm O}(2^{\frac{n-1}{2}}, {\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 7\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
In the cases of signatures $p-q\equiv1, 5\!\!\mod 8$ ($n$ is odd)
\begin{equation}
{\rm G}^{2i1}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm U}(2^{\frac{n-1}{2}}), & \mbox{$(p,q)=(n,0)$;}\\
{\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), & \parbox{.5\linewidth}{$n\equiv 1, 5\!\!\mod 8$, $q\neq 0$;}\\
{\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 3\!\!\mod 8$;}\\
{\rm O}(2^{\frac{n-1}{2}}, {\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 7\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
In the cases of signatures $p-q\equiv1, 5\!\!\mod 8$ ($n$ is odd)
\begin{equation}
{\rm G}^{2i3}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm U}(2^{\frac{n-1}{2}}), & \mbox{$(p,q)=(0,n)$;}\\
{\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), & \parbox{.5\linewidth}{$n\equiv 3, 7\!\!\mod 8$, $p\neq 0$;}\\
{\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 5\!\!\mod 8$;}\\
{\rm O}(2^{\frac{n-1}{2}}, {\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 1\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
In the cases of signatures $p-q\equiv2, 6\!\!\mod 8$ ($n$ is even)
\begin{equation}
{\rm G}^{2}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm U}(2^{\frac{n-2}{2}}), & \mbox{$(p,q)=(0,n), (0,n)$;}\\
{\rm U}(2^{\frac{n-4}{2}},2^{\frac{n-4}{2}}), & \parbox{.5\linewidth}{$n\equiv 2, 6\!\!\mod 8$, $p, q\neq 0$;}\\
{\rm Sp}(2^{\frac{n-4}{2}},{\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 4\!\!\mod 8$;}\\
{\rm O}(2^{\frac{n-2}{2}}, {\Bbb C}), & \parbox{.5\linewidth}{$n\equiv 0\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
\end{thm}
We use the standard notation of classical matrix groups \cite{Cornw}
\begin{eqnarray}
{\rm U}(n)&=&\{A\in{\rm Mat}(n,{\Bbb C}): A^\dagger A={\bf 1}\},\nonumber\\
{\rm U}(p,q)&=&\{A\in{\rm Mat}(p+q,{\Bbb C}): A^\dagger \eta A=\eta\},\nonumber\\
{\rm Sp}(n,{\Bbb C})&=&\{A\in{\rm Mat}(2n,{\Bbb C}): A^T \Omega A=\Omega\},\nonumber\\
{\rm O}(n,{\Bbb C})&=&\{A\in{\rm Mat}(n,{\Bbb C}): A^T A={\bf 1}\},\nonumber
\end{eqnarray}
where $\Omega$ is the block matrix $\Omega=\left( \begin{array}{ll}
 0 & -{\bf1} \\
 {\bf1} & 0 \end{array}\right)$.

\proof We consider Lie group ${\rm G}^{23}_{p,q}$ in Clifford algebra ${C}\!\ell_{p,q}\simeq {\rm Mat}(\frac{n-1}{2},{\Bbb C})$, $p-q\equiv3, 7\!\!\mod 8$. We use complex matrix representation $\beta: U\to \beta(U)$, $U\in{C}\!\ell_{p,q}$ in the cases $p-q\equiv3, 7\!\!\mod 8$ from Section \ref{sectmatr} such that
$\beta(U^\dagger)=(\beta(U))^\dagger$.

In the case of signature $(n,0)$ we have $\tilde{U}=U^\dagger$ (see (\ref{1rt})). From $\tilde{U} U=e$ we obtain $U^\dagger U={\bf 1}$ and isomorphism with group ${\rm U}(2^{\frac{n-1}{2}})$.

Consider the case $q\neq 0$. If $p$ is odd and $q$ is even, then $\tilde{U}=e_{1\ldots p}U^\dagger e^{1\ldots p}$. In the case $p\equiv1\!\!\mod 4$ (in this case we have $q\equiv2\!\!\mod 4$ and $n\equiv3\!\!\mod 4$) let us consider matrix $M=\beta^{1\ldots p}$. We have $M^2=(-1)^{\frac{p(p-1)}{2}}{\bf 1}={\bf 1}$, $M^\dagger=M^{-1}$, ${\rm tr} M=0$. Thus, the spectrum of matrix $M$ consists of the same number of $1$ and $-1$. So there exists such unitary matrix $T^\dagger=T^{-1}$ that $J=T^{-1}MT$, where $J={\rm diag}(1, \ldots, 1, -1, \ldots ,-1)$ is the diagonal matrix with the same number of $1$ and $-1$ on the diagonal. Now we consider transformation $T^{-1}\beta^a T=\gamma^a$ and obtain another matrix representation $\gamma$ of Clifford algebra with $\gamma^{1\ldots p}=J$. From $\tilde{U} U=e$ we obtain $U^\dagger J U = J$ and isomorphism with the group ${\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}})$. The case $p\equiv3\!\!\mod 4$ (in this case we have $q\equiv0\!\!\mod 4$, $n\equiv3\!\!\mod 4$) is similar (we must consider element $M=i \beta^{1\ldots p}$).

Now let $p$ be even and $q$ be odd. Then $n\equiv1\!\!\mod 4$. We know that $k$ is odd and $l$ is even in these cases (see previous section). So we can use formula $U^{\rm T}=e_{b_1 \ldots b_k}\tilde{U} e^{b_1 \ldots b_k}$. Let us consider matrix $M=\beta^{b_1 \ldots b_k}$. We have $M^T=M^{-1}$, ${\rm tr} M=0$, $M^2={\bf 1}$ (or $M^2=-{\bf 1}$). Thus there exists such orthogonal matrix $T^T=T^{-1}$ that $J=T^{-1}MT$ (or $\Omega=T^{-1}MT$, because for $\Omega$ we also have $\Omega^T=\Omega$, $\Omega^2=-{\bf 1}$, ${\rm tr}\Omega=0$).
We consider transformation $T^{-1}\beta^a T=\gamma^a$ and obtain another matrix representation $\gamma$ of Clifford algebra with $\gamma^{b_1\ldots b_k}=J$ (or $\gamma^{b_1\ldots b_k}=\Omega$). So we obtain condition $U^T J U=J$ (or $U^T \Omega U=\Omega$). Thus we obtain in the case $n\equiv 1\!\!\mod 8$ isomorphism with group ${\rm O}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}, {\Bbb C})\simeq{\rm O}(2^{\frac{n-1}{2}},{\Bbb C})$ and in the case $n\equiv 5\!\!\mod 8$ isomorphism with group ${\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C})$. We take into account that $\dim {\rm G}^{23}_{p,q}=2^{n-1}-2^{\frac{n-1}{2}}\sin{\frac{\pi (n+1)}{4}}$, $\dim {\rm O}(2^{\frac{n-1}{2}},{\Bbb C})=2^{n-1}-2^{\frac{n-1}{2}}$, $\dim {\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C})=2^{n-1}+2^{\frac{n-1}{2}}$.

We can obtain isomorphisms for the group ${\rm G}^{12}_{p,q}$ analogously. The statement for the groups ${\rm G}^{2i1}_{p,q}$, ${\rm G}^{2i3}_{p,q}$ and ${\rm G}^{2}_{p,q}$ follows from Theorem \ref{theoremggg}. $\blacksquare$

\begin{thm} We have the following Lie group isomorphisms.
In the cases of signatures $p-q\equiv4, 5, 6\!\!\mod 8$
\begin{equation}
{\rm G}^{23}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm Sp}(2^{\frac{n-2}{2}}), & \mbox{$(p,q)=(n,0)$, $n$ is even;}\\
{\rm Sp}(2^{\frac{n-4}{2}}, 2^{\frac{n-4}{2}}), & \parbox{.5\linewidth}{$n\equiv 4, 6\!\!\mod 8$, $q\neq 0$;}\\
{\rm O}(2^{\frac{n-2}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 0, 2\!\!\mod 8$;}\\
{\rm Sp}(2^{\frac{n-3}{2}})\oplus {\rm Sp}(2^{\frac{n-3}{2}}), & \mbox{$(p,q)=(n,0)$, $n$ is odd;}\\
{\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}})\oplus {\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}}), & \parbox{.5\linewidth}{$n\equiv 5\!\!\mod 8$, $q\neq 0$;}\\
{\rm O}(2^{\frac{n-3}{2}},{\Bbb H})\oplus {\rm O}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 1\!\!\mod 8$;}\\
{\rm GL}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 3, 7\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
In the cases of signatures $p-q\equiv4, 5, 6\!\!\mod 8$
\begin{equation}
{\rm G}^{12}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm Sp}(2^{\frac{n-2}{2}}), & \mbox{$(p,q)=(0,n)$, $n$ is even;}\\
{\rm Sp}(2^{\frac{n-4}{2}}, 2^{\frac{n-4}{2}}), & \parbox{.5\linewidth}{$n\equiv 2, 4\!\!\mod 8$, $p\neq 0$;}\\
{\rm O}(2^{\frac{n-2}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 0, 6\!\!\mod 8$;}\\
{\rm Sp}(2^{\frac{n-3}{2}})\oplus {\rm Sp}(2^{\frac{n-3}{2}}), & \mbox{$(p,q)=(0,n)$, $n$ is odd;}\\
{\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}})\oplus {\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}}), & \parbox{.5\linewidth}{$n\equiv 3\!\!\mod 8$, $p\neq 0$;}\\
{\rm O}(2^{\frac{n-3}{2}},{\Bbb H})\oplus {\rm O}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 7\!\!\mod 8$;}\\
{\rm GL}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 1, 5\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
In the cases of signatures $p-q\equiv2, 3, 4\!\!\mod 8$
\begin{equation}
{\rm G}^{2i1}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm Sp}(2^{\frac{n-2}{2}}), & \mbox{$(p,q)=(n,0)$, $n$ is even;}\\
{\rm Sp}(2^{\frac{n-4}{2}}, 2^{\frac{n-4}{2}}), & \parbox{.5\linewidth}{$n\equiv 2, 4\!\!\mod 8$, $q\neq 0$;}\\
{\rm O}(2^{\frac{n-2}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 0, 6\!\!\mod 8$;}\\
{\rm Sp}(2^{\frac{n-3}{2}})\oplus {\rm Sp}(2^{\frac{n-3}{2}}), & \mbox{$(p,q)=(n,0)$, $n$ is odd;}\\
{\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}})\oplus {\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}}), & \parbox{.5\linewidth}{$n\equiv 3\!\!\mod 8$, $q\neq 0$;}\\
{\rm O}(2^{\frac{n-3}{2}},{\Bbb H})\oplus {\rm O}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 7\!\!\mod 8$;}\\
{\rm GL}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 1, 5\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
In the cases of signatures $p-q\equiv2, 3, 4\!\!\mod 8$
\begin{equation}
{\rm G}^{2i3}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm Sp}(2^{\frac{n-2}{2}}), & \mbox{$(p,q)=(0,n)$, $n$ is even;}\\
{\rm Sp}(2^{\frac{n-4}{2}}, 2^{\frac{n-4}{2}}), & \parbox{.5\linewidth}{$n\equiv 4, 6\!\!\mod 8$, $p\neq 0$;}\\
{\rm O}(2^{\frac{n-2}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 0, 2\!\!\mod 8$;}\\
{\rm Sp}(2^{\frac{n-3}{2}})\oplus {\rm Sp}(2^{\frac{n-3}{2}}), & \mbox{$(p,q)=(0,n)$, $n$ is odd;}\\
{\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}})\oplus {\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}}), & \parbox{.5\linewidth}{$n\equiv 5\!\!\mod 8$, $p\neq 0$;}\\
{\rm O}(2^{\frac{n-3}{2}},{\Bbb H})\oplus {\rm O}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 1\!\!\mod 8$;}\\
{\rm GL}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 3, 7\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
In the cases of signatures $p-q\equiv3, 4, 5\!\!\mod 8$
\begin{equation}
{\rm G}^{2}_{p,q}\simeq\left\lbrace
\begin{array}{ll}
{\rm Sp}(2^{\frac{n-3}{2}}), & \mbox{$(p,q)=(n,0), (0,n)$, $n$ is odd;}\\
{\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}}), & \parbox{.5\linewidth}{$n\equiv 3, 5\!\!\mod 8$, $p, q\neq 0$;}\\
{\rm O}(2^{\frac{n-3}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 1, 7\!\!\mod 8$;}\\
{\rm Sp}(2^{\frac{n-4}{2}})\oplus {\rm Sp}(2^{\frac{n-4}{2}}), & \mbox{$(p,q)=(n,0), (0,n)$, $n$ is even;}\\
{\rm Sp}(2^{\frac{n-6}{2}}, 2^{\frac{n-6}{2}})\oplus {\rm Sp}(2^{\frac{n-6}{2}}, 2^{\frac{n-6}{2}}), & \parbox{.5\linewidth}{$n\equiv 4\!\!\mod 8$, $p, q\neq 0$;}\\
{\rm O}(2^{\frac{n-4}{2}},{\Bbb H})\oplus {\rm O}(2^{\frac{n-4}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 0\!\!\mod 8$;}\\
{\rm GL}(2^{\frac{n-4}{2}},{\Bbb H}), & \parbox{.5\linewidth}{$n\equiv 2, 6\!\!\mod 8$.}
\end{array}
\right.\nonumber
\end{equation}
\end{thm}
We use the standard notation of classical matrix groups \cite{Cornw}
\begin{eqnarray}
{\rm GL}(n,{\Bbb H})&=&\{A\in{\rm Mat}(n,{\Bbb H}): \exists A^{-1} \},\nonumber\\
{\rm Sp}(n)&=&\{A\in{\rm GL}(n,{\Bbb H}): A^*A={\bf 1}\},\nonumber\\
{\rm Sp}(p,q)&=&\{A\in{\rm GL}(p+q,{\Bbb H}): A^* \eta A=\eta\},\nonumber\\
{\rm O}(n,{\Bbb H})&=&O^{*}(2n)=\{A\in{\rm GL}(n,{\Bbb H}): A^* i {\bf 1} A=i {\bf 1}\},\nonumber
\end{eqnarray}
where $*$ is the operation of conjugate transpose of matrix of quaternions.

\proof Let us consider Lie group ${\rm G}^{23}_{p,q}$. In the cases $p-q\equiv4, 6\!\!\mod 8$ we have ${C}\!\ell_{p,q}\simeq {\rm Mat}(2^{\frac{n-2}{2}},\mathbb H)$. In the cases $p-q\equiv5\!\!\mod 8$ we have ${C}\!\ell_{p,q}\simeq {\rm Mat}(2^{\frac{n-3}{2}},\mathbb H)\oplus {\rm Mat}(2^{\frac{n-3}{2}},\mathbb H)$ and we use block-diagonal representation $\beta$ of Clifford algebra (see Section \ref{sectmatr}).

In the case of signature $(n,0)$ we have $U^\dagger=\tilde{U}$ and obtain $U^\dagger U=e$. We have $(\beta(U))^*=\beta(U^\dagger)$, where $*$ is the operation of conjugate transpose of matrices of quaternions. So we obtain isomorphism with the group ${\rm Sp}(2^{\frac{n-2}{2}})$ in the case of even $n$ and isomorphism with the group ${\rm Sp}(2^{\frac{n-3}{2}})\oplus {\rm Sp}(2^{\frac{n-3}{2}})$ in the case of odd $n$.

Now let us consider the case $q\neq 0$. Let $n$ be even. If $p$ and $q$ are even, then we have $U^\dagger=e_{p+1 \ldots n}\tilde{U} e^{p+1 \ldots n}$ and obtain $U^\dagger e^{p+1 \ldots n} U= e^{p+1 \ldots n}$. If $p$ and $q$ are odd, then we have $U^\dagger=e_{1 \ldots p}\tilde{U}e^{1 \ldots p}$
and obtain $U^\dagger e^{1 \ldots p} U= e^{1 \ldots p}$. Let us consider matrix $M=\beta^{p+1 \ldots n}$ (or $\beta^{1\ldots p}$ in the corresponding case). We have $M^*=M^{-1}$ and $M^2={\bf \pm 1}$, so $M^*=\pm M$. It is known (see \cite{quaternion}) that any square quaternionic matrix of size $m$ has the Jordan canonical form with (standard) $m$ right eigenvalues (which are complex numbers with nonnegative imaginary parts) on the diagonal. If matrix is normal ($A^*A=A A^*$), then there exists a unitary matrix $T^*=T^{-1}$ such that $T^* A T$ is a diagonal matrix with standard right eigenvalues on the diagonal. If this  matrix is Hermitian, then these right eigenvalues are real. We obtain that there exists such element $T^*=T^{-1}$ that $\eta=T^{-1}MT$ or $i{\bf 1}=T^{-1}MT$. We use transformation $T^{-1}\beta^a T=\gamma^a$ and obtain another matrix representation $\gamma$ of Clifford algebra with $\gamma^{p+1\ldots n}=\eta$ or $i{\bf 1}$ (or analogously for element $\gamma^{1\ldots p}$). We obtain $ U^* \eta U=\eta$ or $U^* i {\bf 1} U=i {\bf 1}$ and isomorphisms with group ${\rm Sp}(2^{\frac{n-4}{2}}, 2^{\frac{n-4}{2}})$ or ${\rm O}(2^{\frac{n-2}{2}},{\Bbb H})$. We take into account that $\dim {\rm G}^{23}_{p,q}=2^{n-1}-2^{\frac{n-1}{2}}\sin{\frac{\pi (n+1)}{4}}$, $\dim {\rm Sp}(2^{\frac{n-4}{2}},2^{\frac{n-4}{2}})=2^{n-1}+2^{\frac{n-2}{2}}$ and $\dim {\rm O}(2^{\frac{n-2}{2}},{\Bbb H})=2^{n-1}-2^{\frac{n-2}{2}}$.

If $n\equiv 1, 5\!\!\mod 8$, then $p$ is odd and $q$ is even. So we can use formula $U^\dagger=e_{p+1 \ldots n}\tilde{U}e^{p+1 \ldots n}$ and obtain $U^\dagger e^{p+1 \ldots n} U= e^{p+1 \ldots n}$. Our considerations are like considerations in the case of even $n$ but our matrix representation is block-diagonal. We obtain isomorphisms with groups ${\rm O}(2^{\frac{n-3}{2}},{\Bbb H})\oplus {\rm O}(2^{\frac{n-3}{2}},{\Bbb H})$ or ${\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}})\oplus {\rm Sp}(2^{\frac{n-5}{2}}, 2^{\frac{n-5}{2}})$ in different cases.

If $n\equiv 3, 7\!\!\mod 8$, then $p$ is even and $q$ is odd. So, we have $U^*=e_{1\ldots p}\tilde{\hat{U}}e^{1 \ldots p}$ and $\tilde{U}=(e^{1\ldots p})^{-1}\hat{U}^{*}e^{1\ldots p}$. We have block-diagonal matrix representation $\beta$ (see Section \ref{sectmatr}). The even part of an arbitrary Clifford algebra element has the form ${\rm diag}(A,A)$ and the odd part of an element has the form ${\rm diag}(B,-B)$. Then we obtain from $\tilde{U} U=e$:
$$({\rm diag}(A-B, A+B))^* {\rm diag}(G,G){\rm diag}(A+B,A-B)={\rm diag}(G,G),$$
where ${\rm diag}(G,G)$ is matrix representation of element $e^{1\ldots p}$. So, we obtain $(A-B)^T G(A+B)=G$ and isomorphism with group ${\rm GL}(2^{\frac{n-3}{2}},{\Bbb H})$ because for any invertible matrix $A$ there exists such matrix $B$ that $(A-B)^T G(A+B)=G$. Note, that in particular case $p=0$ we have $U^*=\tilde{\hat{U}}$ and we obtain $\hat{U}^{*}U=e$, $({\rm diag}(A-B, A+B))^*{\rm diag}(A+B,A-B)={\bf 1}$, $(A-B)^T(A+B)={\bf 1}$,  and isomorphism with linear group again.

We can obtain isomorphisms for the group ${\rm G}^{12}_{p,q}$ analogously. The statement for the groups ${\rm G}^{2i1}_{p,q}$, ${\rm G}^{2i3}_{p,q}$ and ${\rm G}^{2}_{p,q}$ follows from Theorem \ref{theoremggg}. $\blacksquare$

\section{Relation between group ${\rm G}^{2}_{p,q}$ and spin group}
\label{SectionSpin}

Note that spin group \cite{Lounesto}, \cite{spin}
$${\rm Spin}_{+}(p,q)=\{U\in{C}\!\ell^{(0)}_{p,q} | \forall x\in{C}\!\ell^1_{p,q}, UxU^{-1}\in{C}\!\ell^1_{p,q}, \tilde{U}U=e\}$$
is a subgroup of all five considered Lie groups (\ref{grou}). Moreover, group ${\rm Spin}_{+}(p,q)$ coincides with group
$${\rm G}^{2}_{p,q}=\{U\in{C}\!\ell^{(0)}_{p,q}\, | \, \tilde{U}U=e\}$$
in the cases of dimensions $n\leq 5$. Lie algebra ${C}\!\ell^2_{p,q}$ of Lie group ${\rm Spin}_{+}(p,q)$ is a subalgebra of algebras (\ref{algebr}). Moreover, Lie algebra ${C}\!\ell^2_{p,q}$ coincides with Lie algebra $\overline{\textbf{2}}$ in the cases of dimensions $n\leq 5$, because the notion of grade 2 and the notion of quaternion type 2 coincide in these cases.

Let us represent all isomorphisms of the group ${\rm G}^{2}_{p,q}$ in the following endless table. We use notation $^2{\rm Sp}(1)={\rm Sp}(1)\oplus{\rm Sp}(1)$ and similar notation.

\begin{center}
{\footnotesize
\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|c|c|}
\hline
$p \diagdown q$ & 0       & 1          & 2            &  3          & 4          & 5             & 6            & 7           \\ \hline \hline
0   & ${\rm O}(1)$ & ${\rm O}(1)$    & ${\rm U}(1)$      &  ${\rm Sp}(1)$  & $^2{\rm Sp}(1)$  &   ${\rm Sp}(2)$  & ${\rm U}(4)$      & ${\rm O}(8)$      \\ \hline
1   & ${\rm O}(1)$ & ${\rm GL}(1,{\Bbb R})$& ${\rm Sp}(1,{\Bbb R})$  & ${\rm Sp}(1,{\Bbb C})$ & ${\rm Sp}(1,1)$ & ${\rm GL}(2,{\Bbb H})$& ${\rm O}(4,{\Bbb H})$ & ${\rm O}(8,{\Bbb C})$ \\ \hline
2   & ${\rm U}(1)$ & ${\rm Sp}(1,{\Bbb R})$& $^2{\rm Sp}(1,{\Bbb R})$& ${\rm Sp}(2,{\Bbb R})$ & ${\rm U}(2,2)$  & ${\rm O}(4,{\Bbb H})$   & $^2{\rm O}(4,{\Bbb H})$  & ${\rm O}(8,{\Bbb H})$     \\ \hline
3   & ${\rm Sp}(1)$& ${\rm Sp}(1,{\Bbb C})$& ${\rm Sp}(2,{\Bbb R})$  & ${\rm GL}(4,{\Bbb R})$ & ${\rm O}(4,4)$  & ${\rm O}(8,{\Bbb C})$   &${\rm O}(8,{\Bbb H})$ & ${\rm GL}(8,{\Bbb H})$       \\ \hline
4   &$^2{\rm Sp}(1)$ & ${\rm Sp}(1,1)$ & ${\rm U}(2,2)$    & ${\rm O}(4,4)$   &$^2{\rm O}(4,4)$ & ${\rm O}(8,8)$     & ${\rm U}(8,8)$    & ${\rm Sp}(8,8)$  \\ \hline
5   & ${\rm Sp}(2)$&${\rm GL}(2,{\Bbb H})$ & ${\rm O}(4,{\Bbb H})$  &${\rm O}(8,{\Bbb C})$  & ${\rm O}(8,8)$  & ${\rm GL}(16,{\Bbb R})$  & ${\rm Sp}(16,{\Bbb R})$ &${\rm Sp}(16,{\Bbb C})$     \\ \hline
6   & ${\rm U}(4)$ &${\rm O}(4,{\Bbb H})$ & $^2{\rm O}(4,{\Bbb H})$&  ${\rm O}(8,{\Bbb H})$ & ${\rm U}(8,8)$  &${\rm Sp}(16,{\Bbb R})$   &$^2{\rm Sp}(16,{\Bbb R})$&${\rm Sp}(32,{\Bbb R})$\\ \hline
7   &${\rm O}(8)$  & ${\rm O}(8,{\Bbb C})$&  ${\rm O}(8,{\Bbb H})$   & ${\rm GL}(8,{\Bbb H})$ & ${\rm Sp}(8,8)$ &${\rm Sp}(16,{\Bbb C})$    &${\rm Sp}(32,{\Bbb R})$  &${\rm GL}(64,{\Bbb R})$    \\ \hline
\end{tabular}
\end{center}
}
\end{center}

Note that isomorphisms of the group ${\rm Spin}_{+}(p,q)$ are well-known (see, for example, \cite{Lounesto}, table on the page 224):

\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|c|}
\hline
$p \diagdown q$ & 0 & 1 & 2 & 3 & 4 & 5 & 6   \\ \hline \hline
0 & ${\rm O}(1)$ & ${\rm O}(1)$ & ${\rm U}(1)$ & ${\rm SU}(2)$  & $ ^2{\rm SU}(2)$ & ${\rm Sp}(2)$ & ${\rm SU}(4)$ \\ \hline
1 & ${\rm O}(1)$& ${\rm GL}(1,{\Bbb R})$& ${\rm Sp}(1,{\Bbb R})$ & ${\rm Sp}(1,{\Bbb C})$ & ${\rm Sp}(1,1)$ & ${\rm SL}(2,{\Bbb H})$ &  \\ \hline
2 & ${\rm U}(1)$ & ${\rm Sp}(1,{\Bbb R})$ & $^2{\rm Sp}(1,{\Bbb R})$ & ${\rm Sp}(2,{\Bbb R})$ & ${\rm SU}(2,2)$ &  &   \\ \hline
3 & ${\rm SU}(2)$ & ${\rm Sp}(1,{\Bbb C})$ & ${\rm Sp}(2,{\Bbb R})$ & ${\rm SL}(4,{\Bbb R})$  &  &  &   \\ \hline
4 & $ ^2{\rm SU}(2)$& ${\rm Sp}(1,1)$ & ${\rm SU}(2,2)$ &  &  &  &   \\ \hline
5 & ${\rm Sp}(2)$ & ${\rm SL}(2,{\Bbb H})$ &  &  &  &  &   \\ \hline
6 & ${\rm SU}(4)$ &  &  &  &  &  &   \\ \hline
\end{tabular}
\end{center}

So, in the cases $n\leq 5$ the tables coincide (note, that ${\rm SU}(2)\simeq {\rm Sp}(1)$). In the cases $n=6$ condition $U^{-1}{C}\!\ell^1_{p,q}U\in{C}\!\ell^1_{p,q}$ in the definition of the group ${\rm Spin}_{+}(p,q)$ transforms into condition $\det \gamma(U)=1$ for matrix representation $\gamma$ and we obtain special groups.

Note that group ${\rm Spin}_{+}(p,q)$ in the cases $n\geq 7$ is not directly related to classical matrix groups (see \cite{Lounesto}, p.224).
But we present classical matrix groups that contain this group in the cases of arbitrary dimension $n\geq 7$ and signature $(p,q)$.

\section{Conclusion}\label{sectconcl}

Using results of the present paper and \cite{first}, we can represent isomorphisms between considered five Lie groups and classical matrix groups in the following tables. There is $n\!\!\mod 8$ in the lines and there is $p-q\!\!\mod 8$ in the columns.

$${\rm G}^{12}_{p,q}=\{U\in{C}\!\ell_{p,q}\, | \, \tilde{\hat{U}}U=e\}:$$
\begin{center}
\begin{tabular}{|c||c|c|}
\hline
$n \diagdown p-q$ & $0, 2$              & $4, 6$    \\ \hline \hline
$0, 6$   & $\begin{array}{l} {\rm O}(2^{\frac{n-2}{2}},2^{\frac{n-2}{2}}), p\neq 0;\\ {\rm O}(2^{\frac{n}{2}}), p=0. \end{array}$

         & ${\rm O}(2^{\frac{n-2}{2}},{\Bbb H})$    \\ \hline
$2, 4$   & ${\rm Sp}(2^{\frac{n-2}{2}},{\Bbb R})$

         & $\begin{array}{l} {\rm Sp}(2^{\frac{n-4}{2}},2^{\frac{n-4}{2}}), p\neq 0;\\ {\rm Sp}(2^\frac{n-2}{2}), p=0. \end{array}$       \\ \hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|c||c|c|c|}
\hline
$n \diagdown p-q$ & $1$       & $3, 7$         & $5$    \\ \hline \hline
$7$      & $\begin{array}{l} ^2{\rm O}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), p\neq 0;\\ ^2{\rm O}(2^{\frac{n-1}{2}}), p=0. \end{array}$
         & ${\rm O}(2^{\frac{n-1}{2}},{\Bbb C})$
         & $^2{\rm O}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
$3$      & $^2{\rm Sp}(2^{\frac{n-3}{2}},{\Bbb R})$
         & ${\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C})$
         & $\begin{array}{l} ^2{\rm Sp}(2^{\frac{n-5}{2}},2^{\frac{n-5}{2}}), p\neq 0;\\ ^2{\rm Sp}(2^{\frac{n-3}{2}}), p=0. \end{array}$      \\ \hline
$1, 5$   & ${\rm GL}(2^{\frac{n-1}{2}},{\Bbb R})$
         & $\begin{array}{l} {\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), p\neq 0;\\ {\rm U}(\frac{n-1}{2}), p=0. \end{array}$
         & ${\rm GL}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
\end{tabular}
\end{center}

$${\rm G}^{23}_{p,q}=\{U\in{C}\!\ell_{p,q}\, | \, \tilde{U}U=e\}:$$
\begin{center}
\begin{tabular}{|c||c|c|}
\hline
$n \diagdown p-q$ & $0, 2$              & $4, 6$    \\ \hline \hline
$0, 2$   & $\begin{array}{l} {\rm O}(2^{\frac{n-2}{2}},2^{\frac{n-2}{2}}), q\neq 0;\\ {\rm O}(2^{\frac{n}{2}}), q=0. \end{array}$

         & ${\rm O}(2^{\frac{n-2}{2}},{\Bbb H})$    \\ \hline
$4, 6$   & ${\rm Sp}(2^{\frac{n-2}{2}},{\Bbb R})$

         & $\begin{array}{l} {\rm Sp}(2^{\frac{n-4}{2}},2^{\frac{n-4}{2}}), q\neq 0;\\ {\rm Sp}(2^\frac{n-2}{2}), q=0. \end{array}$       \\ \hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|c||c|c|c|}
\hline
$n \diagdown p-q$ & $1$       & $3, 7$         & $5$    \\ \hline \hline
$1$      & $\begin{array}{l} ^2{\rm O}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), q\neq 0;\\ ^2{\rm O}(2^{\frac{n-1}{2}}), q=0. \end{array}$
         & ${\rm O}(2^{\frac{n-1}{2}},{\Bbb C})$
         & $^2{\rm O}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
$5$      & $^2{\rm Sp}(2^{\frac{n-3}{2}},{\Bbb R})$
         & ${\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C})$
         & $\begin{array}{l} ^2{\rm Sp}(2^{\frac{n-5}{2}},2^{\frac{n-5}{2}}), q\neq 0;\\ ^2{\rm Sp}(2^{\frac{n-3}{2}}), q=0. \end{array}$      \\ \hline
$3, 7$   & ${\rm GL}(2^{\frac{n-1}{2}},{\Bbb R})$
         & $\begin{array}{l} {\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), q\neq 0;\\ {\rm U}(\frac{n-1}{2}), q=0. \end{array}$
         & ${\rm GL}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
\end{tabular}
\end{center}

$${\rm G}^{2i1}_{p,q}=\{U\in{C}\!\ell^{(0)}_{p,q}\oplus i{C}\!\ell^{(1)}_{p,q} : U^\ddagger U=e\}:$$
\begin{center}
\begin{tabular}{|c||c|c|}
\hline
$n \diagdown p-q$ & $0, 6$              & $2, 4$    \\ \hline \hline
$0, 6$   & $\begin{array}{l} {\rm O}(2^{\frac{n-2}{2}},2^{\frac{n-2}{2}}), q\neq 0;\\ {\rm O}(2^{\frac{n}{2}}), q=0. \end{array}$

         & ${\rm O}(2^{\frac{n-2}{2}},{\Bbb H})$    \\ \hline
$2, 4$   & ${\rm Sp}(2^{\frac{n-2}{2}},{\Bbb R})$

         & $\begin{array}{l} {\rm Sp}(2^{\frac{n-4}{2}},2^{\frac{n-4}{2}}), q\neq 0;\\ {\rm Sp}(2^\frac{n-2}{2}), q=0. \end{array}$       \\ \hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|c||c|c|c|}
\hline
$n \diagdown p-q$ & $7$       & $1, 5$         & $3$    \\ \hline \hline
$7$      & $\begin{array}{l} ^2{\rm O}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), q\neq 0;\\ ^2{\rm O}(2^{\frac{n-1}{2}}), q=0. \end{array}$
         & ${\rm O}(2^{\frac{n-1}{2}},{\Bbb C})$
         & $^2{\rm O}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
$3$      & $^2{\rm Sp}(2^{\frac{n-3}{2}},{\Bbb R})$
         & ${\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C})$
         & $\begin{array}{l} ^2{\rm Sp}(2^{\frac{n-5}{2}},2^{\frac{n-5}{2}}), q\neq 0;\\ ^2{\rm Sp}(2^{\frac{n-3}{2}}), q=0. \end{array}$      \\ \hline
$1, 5$   & ${\rm GL}(2^{\frac{n-1}{2}},{\Bbb R})$
         & $\begin{array}{l} {\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), q\neq 0;\\ {\rm U}(\frac{n-1}{2}), q=0. \end{array}$
         & ${\rm GL}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
\end{tabular}
\end{center}

$${\rm G}^{2i3}_{p,q}=\{U\in{C}\!\ell^{(0)}_{p,q}\oplus i{C}\!\ell^{(1)}_{p,q} : \hat{U}^{\ddagger} U=e\}:$$
\begin{center}
\begin{tabular}{|c||c|c|}
\hline
$n \diagdown p-q$ & $0, 6$              & $2, 4$    \\ \hline \hline
$0, 2$   & $\begin{array}{l} {\rm O}(2^{\frac{n-2}{2}},2^{\frac{n-2}{2}}), p\neq 0;\\ {\rm O}(2^{\frac{n}{2}}), p=0. \end{array}$

         & ${\rm O}(2^{\frac{n-2}{2}},{\Bbb H})$    \\ \hline
$4, 6$   & ${\rm Sp}(2^{\frac{n-2}{2}},{\Bbb R})$

         & $\begin{array}{l} {\rm Sp}(2^{\frac{n-4}{2}},2^{\frac{n-4}{2}}), p\neq 0;\\ {\rm Sp}(2^\frac{n-2}{2}), p=0. \end{array}$       \\ \hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|c||c|c|c|}
\hline
$n \diagdown p-q$ & $7$       & $1, 5$         & $3$    \\ \hline \hline
$1$      & $\begin{array}{l} ^2{\rm O}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), p\neq 0;\\ ^2{\rm O}(2^{\frac{n-1}{2}}), p=0. \end{array}$
         & ${\rm O}(2^{\frac{n-1}{2}},{\Bbb C})$
         & $^2{\rm O}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
$5$      & $^2{\rm Sp}(2^{\frac{n-3}{2}},{\Bbb R})$
         & ${\rm Sp}(2^{\frac{n-3}{2}},{\Bbb C})$
         & $\begin{array}{l} ^2{\rm Sp}(2^{\frac{n-5}{2}},2^{\frac{n-5}{2}}), p\neq 0;\\ ^2{\rm Sp}(2^{\frac{n-3}{2}}), p=0. \end{array}$      \\ \hline
$3, 7$   & ${\rm GL}(2^{\frac{n-1}{2}},{\Bbb R})$
         & $\begin{array}{l} {\rm U}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), p\neq 0;\\ {\rm U}(\frac{n-1}{2}),p=0. \end{array}$
         & ${\rm GL}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
\end{tabular}
\end{center}

$${\rm G}^{2}_{p,q}=\{U\in{C}\!\ell^{(0)}_{p,q}\, | \, \tilde{U}U=e\}:$$
\begin{center}
\begin{tabular}{|c||c|c|}
\hline
$n \diagdown p-q$ & $1, 7$              & $3, 5$    \\ \hline \hline
$1, 7$   & $\begin{array}{l} {\rm O}(2^{\frac{n-3}{2}},2^{\frac{n-3}{2}}), p, q\neq 0;\\ {\rm O}(2^{\frac{n-1}{2}}), (n,0), (0,n). \end{array}$

         & ${\rm O}(2^{\frac{n-3}{2}},{\Bbb H})$    \\ \hline
$3, 5$   & ${\rm Sp}(2^{\frac{n-3}{2}},{\Bbb R})$

         & $\begin{array}{l} {\rm Sp}(2^{\frac{n-5}{2}},2^{\frac{n-5}{2}}), p, q\neq 0;\\ {\rm Sp}(2^\frac{n-3}{2}), (n,0), (0,n). \end{array}$       \\ \hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{|c||c|c|c|}
\hline
$n \diagdown p-q$ & $0$       & $2, 6$         & $4$    \\ \hline \hline
$0$      & $\begin{array}{l} ^2{\rm O}(2^{\frac{n-4}{2}},2^{\frac{n-4}{2}}), p, q\neq 0;\\ ^2{\rm O}(2^{\frac{n-2}{2}}), (n,0), (0,n). \end{array}$
         & ${\rm O}(2^{\frac{n-2}{2}},{\Bbb C})$
         & $^2{\rm O}(2^{\frac{n-4}{2}},{\Bbb H})$    \\ \hline
$4$      & $^2{\rm Sp}(2^{\frac{n-4}{2}},{\Bbb R})$
         & ${\rm Sp}(2^{\frac{n-4}{2}},{\Bbb C})$
         & $\begin{array}{l} ^2{\rm Sp}(2^{\frac{n-6}{2}},2^{\frac{n-6}{2}}), p, q\neq 0;\\ ^2{\rm Sp}(2^{\frac{n-4}{2}}), (n,0), (0,n). \end{array}$      \\ \hline
$2, 6$   & ${\rm GL}(2^{\frac{n-2}{2}},{\Bbb R})$
         & $\begin{array}{l} {\rm U}(2^{\frac{n-4}{2}},2^{\frac{n-4}{2}}), p, q\neq 0;\\ {\rm U}(\frac{n-2}{2}), (n,0), (0,n). \end{array}$
         & ${\rm GL}(2^{\frac{n-4}{2}},{\Bbb H})$    \\ \hline
\end{tabular}
\end{center}

\medskip

Note that if we know isomorphisms between these Lie groups, then we know isomorphisms between corresponding Lie algebras.
So, we also obtain isomorphisms between Lie algebras $\overline{\textbf{2}}\oplus i\overline{\textbf{1}}$, $\overline{\textbf{2}}\oplus i\overline{\textbf{3}}$, $ \overline{\textbf{2}}\oplus\overline{\textbf{3}}$, $\overline{\textbf{2}}\oplus\overline{\textbf{1}}$, $\overline{\textbf{2}}$ and corresponding classical matrix Lie algebras: linear ${\rm gl}(k,{\Bbb R})$, ${\rm gl}(k,{\Bbb H})$, unitary ${\rm u}(k)$, ${\rm u}(r,s)$, orthogonal ${\rm so}(k,{\Bbb R})$, ${\rm so}(r,s)$, ${\rm so}(k,{\Bbb C})$, ${\rm so}(k,{\Bbb H})$, symplectic ${\rm sp}(k)$, ${\rm sp}(r,s)$, ${\rm sp}(k,{\Bbb R})$, ${\rm sp}(k,{\Bbb C})$ or direct sums of such Lie algebras of corresponding dimensions in different cases.

As we have already mentioned in the Introduction, group $G^\epsilon_{p,q}$ \cite{Abl2} coincides with the group ${\rm G}^{23}_{p,q}$ in the case of signature $(n,0$) and with the group ${\rm G}^{12}_{p,q}$ in the case of signature $(0,n)$. In \cite{Lounesto} (p. 236, Tables 1 and 2) one finds isomorphisms between classical matrix Lie groups and the groups ${\rm G}^{12}_{p,q}$ and ${\rm G}^{23}_{p,q}$ which are automorphism groups of the scalar products on the spinor spaces. In the present paper we use another technique. We obtain the same isomorphisms for groups ${\rm G}^{12}_{p,q}$ and ${\rm G}^{23}_{p,q}$ and also isomorphisms for groups ${\rm G}^{2i1}_{p,q}$, ${\rm G}^{2i3}_{p,q}$, ${\rm G}^{2}_{p,q}$ using the notion of additional signature and relations between operations of conjugation in Clifford algebra and the corresponding matrix operations. We also study the corresponding Lie algebras which are related to subspaces of quaternion types.

redpajama_set_name: RedPajamaArXiv


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text:
Q: Laravel - Adding back slash to array after edit this is my controller:
$sql = Raid::findOrFail($request['id']);
$sql = $sql->update($request->all());

I have a array in my table , after update value will be like this:
"{\"Plate\":0,\"Cloth\":0,\"Mail\":0,\"Leather\":0}"

but it should be:
{"Plate":"0","Cloth":"0","Mail":"0","Leather":"0"}

so I will get an error
before this , I was updating like this and it was ok:
$sql = Raid::where('id', $request['id'])->update($request->all());

and this is my model (traders and class_traders is fields that I have problem with):
    use SoftDeletes;

    use \OwenIt\Auditing\Auditable;

    protected $table = 'raid';

    protected $dates = ['date_and_time','deleted_at'];

    protected $fillable = [
      'admin_id', '....
    ];

    protected $casts = [
        'bosses' => 'array',
        'traders' => 'array',
        'class_traders' => 'array',
        'boosters' => 'array',
    ];

A: I think what you need to do is don't update the model using the update($request->all()) way, but you need to specify each of the field's value. For the JSON field, you could try using json_encode() to save it.
That should do it.

redpajama_set_name: RedPajamaStackExchange


Combining contemporary and nation home plans can mean combining rustic materials reminiscent of slate flooring, exposed ceiling beams, and kitchens with modern chrome steel home equipment , plastic laminated cupboards , glass shelving and butcher block countertops. The ultimate result is a contemporary kitchen which is serviceable and has a comfortable , rustic feel. Modern design is about smooth surfaces corresponding to glass and stainless steel , and the use of bold colors.
Before we go right down to the actual theme of contemporary house plans, you need to know the essential features of a modern household. For starters up to date house plan has massive home windows to provide a light-weight and comfortable environment, excessive ceilings, versatile and continuous floor plan to accommodate modern furnishings and fixtures; and utilization of modern supplies, resembling glass, metal , vinyl, stone, marble, and so on.
Home designs are highly effective symbols that you should use to create an announcement to the world on who you are. Buildings have a long lifespan and will proceed communicating your assertion to the world lengthy after your departure. House designs are also a mark of self-actualization.
Mid-Century Modern house plans are rising in popularity from New York to LA and in every single place in between. These plans embrace historic Eichler designs from the 1960s, in addition to latest home plans inspired by the iconic `Case Study ` modern houses in Los Angeles of the late Forties and early 1950s. Led by Dwell journal , the mid century aesthetic of open plans, giant windows and minimal detailing is rising as one of the key design developments of the early 21st century.


Rusty's first win in NASCAR happened at a track in California at the stroke of midnight and his favorite car was named Midnight, so we had a pretty good head start on where we were going to go with the brand story. LAVIDGE partnered with Monomyth on the design/story development and LAVIDGE CCO Bob Case created the illustrations. The brand story focused on the magical things that happen at midnight (Paul Revere's ride, midnight mass, Cinderella's night, etc.). To enhance the magical feeling, we created a midnight witness: the Raven (think Edgar Allen Poe).
The project stalled at the finish line, but the work is something we're very proud of. The illustrations were awarded local and regional Addy awards and the illustration of the raven led to Bob Case being selected as one of Lüerzer's Archive Magazine's 200 Best Illustrators Worldwide for 2016-17.


The Independent National Electoral Commission (INEC) says 400,000 Permanent Voters Cards (PVCs) were yet to be collected out of the 3,215, 273 registered in the state.
He said that about 96,735 cards were yet to be received from the commission's headquarters.
"Collection of PVCs is still ongoing in the state at the various LGAs,'' he said.


Unwrap the food and beverage space across social video platforms with Pixability's latest report.
Food and beverage consumers don't limit their online video viewing to a single platform — they're connecting with like-minded audiences across the walled garden ecosystem, comparing products, finding recipes, and making brand and product decisions. Pixability's latest report unwraps the food and beverage ecosystem across YouTube, Facebook, Instagram, Twitter, and Snapchat, to help marketers better shape their video strategies to reach and engage the consumers that matter most to them.
Whether viewers are gourmets, fast food fanatics, or have a sweet tooth, they'll find the video content they're looking for across YouTube, Facebook, Instagram, Twitter, and Snapchat.
Download the highlights of Pixability's food and beverage study now. To get the full report, set up a meeting today.


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Legendary Model Chronicles How She Snuck into the US to Pursue Fashion Career
Kelli Serio
Somalia-born fashion model Iman, born Iman Mohamed Abdulmajid, revealed to New York magazine Tuesday how she snuck into the United States to break into the fashion world more than four decades ago.
The supermodel, who has walked for legendary designers Versace, Calvin Klein, and Donna Karan, said she faced several barriers when trying to launch her career. One of those was traveling to the United States.
"I was under 18, and to leave Kenya to come to the United States, to get a passport, you had to be 18," she said.
Iman was studying at the University of Nairobi at the time, which clashed with her lifelong dream of becoming a supermodel.
"So I lied and said I was 19 to get the passport, because [otherwise] I had to have permission from my parents, and my parents would never have let me come," she continued.
Born to a diplomat and gynecologist, Iman's interest in high fashion presented a challenge within her family. All obstacles aside, the legendary model was still able to fly under the radar and travel to New York City to pursue her dreams.
"You know how they found out? I was in Newsweek a week later," she explained. "It's the only magazine my father reads, and the next thing, he's reading, and there's a picture of me in New York, [saying,] 'She's the hottest model now.'"
Iman, now 59, married David Bowie in 1992. She admits her decision was risky, but one that she does not regret.
"I broke the law!" she said. "But nobody can touch me now, I'm 60 years old, it's too late. The time has passed."
EntertainmentNational SecurityBig HollywoodFashionmodelSomalia

redpajama_set_name: RedPajamaCommonCrawl


My son & nephew started almost 2 months ago and I have nothing but AMAZING things to say about this daycare. Mrs. Sara, Mrs. Kemisha, Mrs. Rhonda, and Mrs. Ally have been AMAZING!!! We were super nervous about putting the babies in daycare and now they always wanna be at school! I wanna thank this daycare from the bottom of my heart for loving our babies the way we do!! Y'all are a blessing!!!!
Excellent school and daycare my little girl loves it there. The teachers are friendly and helpful. Overall great place.
Bearfoot has always treated my girls very good. They have a clean facility and a great management team in place. If you want a safe and fun learning environment, I recommend taking your kids there.
Also, the pricing is reasonable for your age appropriate child.
My daughter started Bearfoot at 7mths, and is about to turn 1! We have had a great experience, and she loves going to "school". The Staff is super friendly and attentive. They always answer any questions and listen to any special instructions or concerns. I feel really good about leaving her and do not worry while I'm at work. Such a relief to know she is happy and in good hands.
I love this daycare!!! I have had my baby in there since she was 6 weeks old. The employees are dependable and trusting. I have a sense of calmness when I drop off and pick up. My older child has done 3 summers there and I wouldn't go anywhere else!!!
My daughter has attended Bearfoot since she was 12 weeks old. Such a loving and caring staff that has taken amazing care of my child as she has grown. She loves learning and going to school. They encourage creativity and she has really excelled in Pre K this year in the McGraw-Hill program.
Really really excellent daycare. Our 3 year old was at Bearfoot since 8 weeks old and our 11 month old since 12 weeks old. We moved to Denver recently and really miss the school and all the teachers. The management at the school is very accommodating and flexible with family situations. The teachers are all very sweet and caring and genuinely love the kids. The facility is clean and has nice rooms and an awesome playground. We wish we had a Bearfoot here in Denver because we haven't found any daycare on par with Bearfoot. We miss you!
This is just a quick note to let you know how very pleased I am with your Private Kindergarten class my son Teige attends. Ms. Heidi is so sweet and Teige really loves coming to school and going to her class. I have also noticed just how much he is learning and I feel so confident on how prepared he will be going to Kindergarten next year. As a parent who has been bringing both of her children here for the past 5 years, I wanted to say thank you for all that Bearfoot does for my children.


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Q: Laravel get Model siblings and self - is this the right way? I have an Eloquent model which has these functions:
<?php

class Package extends Model
{

    public function parent()
    {
        return $this->belongsTo( Package::class, 'parent_id' );
    }

    public function childs()
    {
        return $this->hasMany( Package::class, 'parent_id' );
    }

    /**
     * All packages that belongs to each other
     */
    public function siblings_and_self()
    {
        $result = $this->newCollection()->add( $this );
        $result = $result->merge( $this->parent()->get() );
        $result = $result->merge( $this->childs()->get() );

        return $result;
    }
}

It is about the last function siblings_and_self and I am curious if this is the correct way to retrieve the result.
Or would you suggest another method retrieving all related records and self?

A: Depends on what you are willing to do with the data, but you could use eager loading to make your query look nicer.
$package = Package::where('id', 5)->with(['parent', 'childs'])->get();

Now you can use the parent and child properties to find the related data
// The id of the current package
echo $package->id;

// The id of the parent package
echo $package->parent->id;

// The id of the first child package
echo $package->childs->first()->id;

A: I think it should be improved.
Instead of:
$result->merge( $this->parent()->get() );

you should rather use:
$result->merge([$this->parent]);

Also instead of
$this->childs()->get()

you can use:
$this->childs

So finally you can simplify it and use just:
public function siblings_and_self()
{
    return $this->newCollection([$this, $this->parent])->merge( $this->childs );
}

However if case you want to use this on multiple model, you should first eager load parent and childs relationships to avoid n+1 queries problem like this:
// here you eager load
$packages = Package::with('parent', 'childs')->get();

// and now you can use it
foreach ($packages as $package)
{
   $siblings_and_self = $package->siblings_and_self();
  // and now you can do something with this variable
}


This is another generic integration implemented in The Customer Collector. This one allows you to send the contact data to services using the HTML code of a subscribe form.
There is one caveat, though: this only works with autoresponders allowing frequent submissions originated from the same IP address. For that reason, AWeber, for example, is not suitable at all, because AWeber only allows one submission/IP address/day, and in the context of The Customer Collector, all submission requests would originate from the web server on which it is installed.
A comprehensive list of suitable services will be published here soon.
This integration doesn't require prior configuration, so it won't appear on the Connections page. It will only be available as a target connector in the mapping edit form.
In the image above you can see the target portion of Mapping Edit form when "Subscribe via Webform" is selected as the target. In order to configure the mapping, you first need to paste the HTML code of the form to be used. Then the form fields will appear in the 3 dropdown lists, allowing you to decide which of them will receive which information. If in doubt, please consult the documentation of the respective autoresponder to understand its form field naming convention.


A very confusing game, Confusion. There are 3 dice; Color dice, Word dice and an Action dice. They each have a background color and some other piece of information. If the word on the word dice and the color on the color dice don't match be the first to throw out a card of the color that is missing (there are only 3 colors total). If they do match proform the action on the action dice. Collect the most points by not being last or freezing up. There is also a variant that complicates things further.


Type treatments and custom letterforms for various branding and promotional jobs.
Clients looking to stake out a new place in their chosen market are always wanting to stand out, and creating a new brand and marketing identification is a great place to start.
Some of these clients have a strong idea of where they wanted to take their brand, and others request our art direction (either of these methods works for us). The branding is usually the first point of reference for the client's audience and - depending on the client - needs to convey the service they provide, who they want to reach, and the industry they work within.
A mood/style board is to be provided or created by us to discover the right direction, and from here we are able to build a collection of thumbnail sketches and gradually zone in on the final direction before final rendering.
Use the existing company ethos or mood to carve out unique and vibrant marks that instantly sell the brand concept.
As we complete the branding we take note of the styles and concepts that were not used in the final id but serve as a guide for complementary assets, and from here we develop type treatments and layouts for marketing materials. This keeps the style of the branding and the marketing harmonious and consistent throughout.


For anyone in leadership, conflict is inevitable. It will happen. And in some seasons it will feel like conflict management is the only thing you do. It could be a struggle over direction, a dispute between staff, or a disagreement with peers, customers, or suppliers. Since conflict is unavoidable, those who anticipate, lead, and successfully manage these moments have a distinct leadership advantage. The next time you encounter conflict, consider these five laws.
Conflict is not always bad, but we can become quickly bothered by it. This is part of the reason for increasing levels of leadership stress. We need to keep in mind conflict is normal. Often it invites clarity where there is confusion and resolves a silent tension that needs to be made known. Yet many leaders conflict-averse because they think conflict is a reflection on them and their leadership deficiencies. But it's not the conflict that's the issue, but rather how they lead through the conflict is more the issue. If as a leader you can determine the root issues for the conflict without getting hooked by emotional insecurities you will always be able to lead more successfully through these encounters. Don't let yourself get baited and hooked. It's hard to help others who are anxious when you are too.
When emotions get elevated, people tend to stop listening to each other because their ability to listen is being hijacked by their emotional need to be heard. You can de-escalate a conflict by sitting down and listening – with precision I might add – and clarifying what is being said. Most of the time in a conflict people want to be both heard and understood. In the process of listening, you are helping them to gain a voice and a hearing they feel like they don't have, and in the process, you get a better understanding of what's happening in your organization. And in your listening give them time to ramp down emotionally.
As we listen, we should be on a quest for identifying the outcomes desired. Sometimes, those who conflict with each other don't realize that they may have similar or even common goals but often just arguing over the best way to get there. If you know the result they both want then you can help them figure out how they can get there - and together. But remember it's not only about the outcomes they want, but it's also about the lessons learned through conflict and the relational wins post-conflict.
Is your next step another meeting with additional people? Is it a change in process or procedure? Is it seeking a new way of doing things? Whatever it is, identify it and ensure that those responsible are clear on what they need to do and when they need to do it. Set accountability in place to ensure that promises and decisions are kept to avoid another eruption. These steps are proof to all parties that you have listened and want to reconcile future issues that will help others manage conflicts better themselves.
Sometimes you have parties who are unwilling to work toward a solution, or the solution involves parties or processes that you don't have the authority to amend or modify. Thus, you need to ask someone to help resolve the matter. Sometimes a human resource department needs to be consulted or authorities with the power to make the necessary changes you cannot. Don't be afraid to ask others when this happens; it's better than hiding the issue or attempting to handle it on your own.
Remember that emotions and insecurities can prevent people from working through issues that are usually simple to solve. If you as a leader can help lower the emotions of the parties and help them look objectively at the items from the perspective of the desired outcome, solutions are often easy to find. Conflict is a healthy test to embrace it as such.


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Review: Tabu Wave 2 – Alexander O'Neal, Cherrelle, Kathy Mathis and The S.O.S. Band
The earth has music for those who listen, proclaimed Clarence Avant's Tabu Records label. A major force in contemporary R&B from the late 1970s through the 1990s, Tabu followed in the footsteps of other black-owned, independent music empires as Berry Gordy's Motown and Kenny Gamble and Leon Huff's Philadelphia International. While Tabu never achieved the same level of crossover success as those aforementioned labels, it indeed picked up the torch of "The Sound of Young America," and its cutting-edge electronic style still reverberates in R&B and hip-hop today. Earlier in 2013, the U.K.'s Demon Music Group announced the reactivation of the Tabu label for an ambitious reissue program, the second wave of which has recently arrived in stores. This batch includes two classic titles from the era-defining production team of Jimmy Jam and Terry Lewis, Alexander O'Neal's Hearsay (1987) and Cherrelle's High Priority (1985), both reissued as double-disc sets. In addition, this wave includes single-disc expansions of The S.O.S. Band's S.O.S. (1980) and Kathy Mathis' Katt Walk (1987).
1980's gold-selling S.O.S. marked the LP debut of The S.O.S. Band: Jason "T.C." Bryant on keyboards and vocals, Billy "B.E." Ellis on saxophone, keyboards and vocals, Mary Davis on vocals and percussion, James Earl Jones III on drums and vocals, Willie "Sonny" Killebrew on saxophone, flute and vocals, Bruno Speight on lead guitar and John Simpson on bass. The album was produced by Sigidi Abdullah; the band's hits with Jam and Lewis would come later. Abdullah co-wrote "Take Your Time (Do It Right)," the No. 1 R&B/No. 12 Pop song that became the band's first calling card. But it's just one of eight essential tracks on this debut album which owes as much to the sound of the seventies as to the new decade it welcomed.
The S.O.S. Band successfully managed to synthesize the effusive but commercially-waning sound of disco with a solid bed of funk and a key pop sensibility that seemingly owed much to Earth, Wind and Fire. Like that group, the S.O.S. Band prominently employed horns for a style that would attract soul fans both young and old, and boasted talented vocalists, including the big-voiced Mary Davis. S.O.S. was truly the organic sound of a band at work, and showed off each side of its style, from slow-burning ballads to catchy dancefloor anthems. Almost every track on the album could have been pulled as a single,
The sleek EWF sound is most evident on "Open Letter," while wistful, Bacharach-esque horns add dimension to the melancholy "What's Wrong with Our Love Affair." The exuberantly up-tempo "Love Won't Wait for Love" emphasizes disco flourishes with its big strings, horns and suitable-for-dancing break. Of course, the sexy smash "Take Your Time (Do It Right)" also had its roots in disco, and hasn't lost any of its luster in the ensuing years. Neither has "Take Love Where You Find It," another big, brassy, disco-flavored track with tasty flute so redolent of the era. Six bonus tracks have been added to S.O.S., including the single edit and disco mix of "S.O.S. (Dit Dit Dit Dash Dash Dash Dit Dit Dit)," the long version and both sides of the single of "Take Your Time (Do It Right)," and the edit of "What's Wrong with Our Love Affair." Justin "Musicology" Kantor provides the liner notes, which contain fresh quotes from Mary Davis and trumpeter Abdul Ra'oof.
After the jump: the scoop on Cherrelle, Alexander O'Neal and Kathy Mathis! Read the rest of this entry »
Posted in Alexander O'Neal, Cherrelle, Kathy Mathis, News, Reissues, Reviews, The S.O.S. Band


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Home > Fiber Optic Cable, Testers, & Hardware > Pre -Terminated Fiber Optic Cable Assemblies > Ultra Thin Indoor Armored Pre-Terminated Fiber Optic Assemblies
12 Strand Custom Indoor Ultra Thin Micro Armor (Plenum) 62.5/125 OM1 Multimode Assembly by QuickTreX
QT-MA-I-M1-12
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Our Ultra-Thin Armored Pre-Terminated Fiber Optic Assemblies utilize Micro Armor Fiber™. It's a revolutionary designed fiber optic cable that provides a perfect solution for your fiber optic installs and usage. Instead of a traditional interlocking armor it utilizes a steel spiraling technology. This allows for the cable to remain extremely flexible while also being crush and rodent resistant. Micro Armor Fiber™ features the smallest outer diameter of any armor compared to conventional fiber optic cable in size and flexibility. Its light weight and great flexibility makes pulling and installs faster and easier. These assemblies are typically up to 65 percent smaller in diameter and 75 percent lighter than the usual aluminum interlocking armor cable. These assemblies provide a perfect combination of strong armor and maximum bend radius. They are available in indoor and outdoor options. They come available in OM1, OM2, OM3, OM4 and Singlemode and we can provide any standard fiber optic connector type such as LC, ST, and SC.
• Made to exacting specifications.
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• Factory terminations and polishing are
superior to virtually all field terminations
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• Optional "Pulling Eyes" (pulling baskets) allow the ends of the
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Micro Armor vs Interlocking Armor
6 Strand Custom Indoor Ultra Thin Micro Armor (Plenum) 62.5/125 OM1 Multimode Assembly by QuickTreX


Professor Lai's research focuses on nanoparticle-based delivery of bioactive molecules to mucosal tissues.
Stephen Mayfield's research analyses the structures and proteins in algae.
Lynn Lewis is a professor of biology at the University of Mary Washington.
Robert Lanza is currently Chief Scientific Officer at Ocata Therapeutics and Adjunct Professor at Wake Forest University School of Medicine.


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Field Automation Leader ProntoForms Processed 7 Million Customer Forms in 2014
V.PFM | January 22, 2015
OTTAWA, Jan. 22, 2014 /CNW/ - ProntoForms Corporation (TSXV: PFM), maker of ProntoForms, the leading mobile data collection solution for smartphones and tablets, is pleased to announce that it processed approximately 7 million customer forms in 2014. At an average of two pages per form processed, ProntoForms helped to reduce the use and filing of 14 million paper sheets and 70 tons of paper in one calendar year.
"We are extremely proud of this form processing achievement. This high data record volume can be attributed not only to the multiple field process use of our product by business customers, but also to its rapidly growing popularity among large enterprises with a high number of workers in the field. We would like to thank our customers for their continued loyalty and we look forward to providing them with even more time-saving, high productivity features in 2015," said Alvaro Pombo, ProntoForms Founder and CEO.
For more information, visit www.prontoforms.com
About ProntoForms® and ProntoForms Corporation
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Matthew Ross, Media Manager, mross@prontoforms.comCopyright CNW Group 2015
January 22, 2015 - 8:59 AM EST
SOFTWARE-APPLICATION TECHNOLOGY


This month, Jan Wedekind, will be describing game development with Ruby.
Using the Ruby programming language, the CWiid library, and the Ruby bindings for OpenGL, it is possible to develop a 3D game which is controlled using a Nintendo Wii Remote. By furthermore invoking RMagick and the ALSA library, one can add textures and sound to the game. Note that the library for accessing the Wii Remote (CWiid) and the sound library (ALSA) require Linux. However it is conceivable to do a similar project with alternate libraries on another operating system.
Someone will be around at the GIST Lab from no later than 6.30pm. The presentation and questions will run 7pm-8.30pm. The GIST Lab is on Grinder's Hill, at one end of the Workstation.


text:
Q: For $a,b,c$ positive real numbers can it be true that: $(ab+bc+ca)^3 \ge (a^2+2b^2)(b^2+2c^2)(c^2+2a^2)$ For $a,b,c$ positive real numbers can it be true that: $$(ab+bc+ca)^3 \ge (a^2+2b^2)(b^2+2c^2)(c^2+2a^2)$$
It really seems unlikely because it reminds me the rearrangement inequality when the direction is flipped.

A: $\prod\limits_{cyc}(a^2+2b^2)\geq(ab+ac+bc)^3\Leftrightarrow\sum\limits_{cyc}(2a^4b^2+4a^4c^2-a^3b^3-3a^3b^2c-3a^3c^2b+a^2b^2c^2)\geq0$,
which is true by Schur and AM-GM:
$\sum\limits_{cyc}(2a^4b^2+4a^4c^2-a^3b^3-3a^3b^2c-3a^3c^2b+a^2b^2c^2)\geq$
$\geq\sum\limits_{cyc}(2a^4c^2+3a^3b^3-3a^3b^2c-3a^3c^2b+a^2b^2c^2)=$
$=3\sum\limits_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)+2\sum\limits_{cyc}(a^4c^2-a^2b^2c^2)\geq0$.


text:
Q: App Invites is deprecated I am developing one application which has appinvite functionality but after pod update i got below below error

'show(from:with:delegate:)' is deprecated: App Invites no longer supported

After that i goto the facebook developer site and found below text on top

With the release of the Facebook SDK version 4.28.0, App Invites is deprecated. It will be supported until February 5, 2018. (link)

Is there any other way to invite on facebook after app invite stop support for non game application in ios?

A: Chirag,
Once app invites are deprecated there is no other way to invite on Facebook unless you allow sharing on users' FB timeline with tracking/attribution links.
As an app developer, you should also consider using popular messaging apps and other social networks to leverage app invites i.e Whatsapp, Kik, Twitter, Kakao etc. More details here
(Disclaimer: I am the founder of GetSocial)