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| \begin{document}
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| %-------------------------------------------------------------------------
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| % editorial commands: to be inserted by the editorial office
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| %\firstpage{1} \volume{228} \Copyrightyear{2004} \DOI{003-0001}
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| %\seriesextraline{This is the Concrete Title of this Book\br H.E. R and S.T.C. W, Eds.}
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| %\submitted{March 14, 2003}
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| %Insert here the title, affiliations and abstract:
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|
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| \title{A hyperbolic non distributive algebra in $1+2$ dimensions}
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| \maketitle
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| \begin{abstract}
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| We introduce a non distributive algebra over the reals in dimensions
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| 1 + 2 that in the special 1 + 1 case recovers the hyperbolic complex
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| numbers. The algebra contains divisors of zero that can be avoided
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| by introducing the necessary conditions. Under these conditions, the
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| proposed addition and product operations satisfy group properties.
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| More stringent conditions sufficient to satisfy group properties separate
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| the algebra in two subspaces. As an example, the composition of velocities
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| in a deformed Lorentz metric is presented. In this approach, Minkowski
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| light cones are deformed into light pyramids.
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| \end{abstract}
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| %%% ----------------------------------------------------------------------
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|
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|
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| \section{Introduction}
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|
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| Systems of hypercomplex numbers, such as quaternions or the octonions
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| are algebras over the real numbers that lack some properties for their
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| product, either commutativity or associativity. %%
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| The search for higher dimensional algebras retaining some of these
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| properties is severely limited by Frobenius and Hurwitz theorems.
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| The former theorem establishes that finite-dimensional associative
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| division algebras over real numbers must be isomorphic to the real
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| $\mathbb{R}$, complex $\mathbb{C}$ or quaternion $\mathbb{H}$ algebras
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| with dimensions 1, 2, and 4, respectively. The latter theorem states
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| that the only normed division algebras over $\mathbb{R}$ are the
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| real $\mathbb{R}$, complex $\mathbb{C}$, quaternion $\mathbb{H}$
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| and the octonions $\mathbb{O}$. Of these algebras quaternions are
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| no longer commutative and octonions are neither commutative nor associative.
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|
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| These theorems require the absence of zero divisors. If distributivity
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| and associativity are fulfilled then the algebra (except for reals
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| and complex) is either not commutative or has divisors of zero. A
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| theorem due to Scheffers (1893) establishes that for distributive
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| systems with unity, the differential calculus does exist only if the
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| systems are commutative \cite{catoni2008}. Differential calculus
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| is crucial in order to define functions of a hypercomplex variable
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| and to associate them with the infinite-dimensional Lie group of functional
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| mappings. For this reason there has been renewed interest in commutative
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| algebras such as Segre's commutative quaternions although they possess
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| zero divisors \cite{catoni2006_1}.
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|
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| Two dimensional hypercomplex numbers contain a real component and
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| an imaginary component. Depending on the product definition of imaginary
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| unit, these hypercomplex numbers are elliptic, parabolic or hyperbolic.
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| Elliptic hypercomplex numbers are the complex numbers, a commutative
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| division algebra. Hyperbolic hypercomplex two dimensional numbers
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| or split-complex numbers form an associative and commutative algebra
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| that is tailor made to describe Minkowski space-time special relativity
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| in $1+1$ dimensions. Higher $N$ dimensional hyperbolic numbers have
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| been explored using direct product rings with zero divisors as a basis
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| \cite{fjelstad+gal1998}.
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|
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| In this paper we introduce an algebra in $1+2$ dimensions that degenerates
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| to hyperbolic $\mathbb{H}_{2}$ numbers when only one director component
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| is present. This algebra, that we call \emph{hyperbolic scator} or
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| \emph{real scator} algebra, and we will see that its product is commutative
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| and associative. However, as we shall see, the product does not distribute
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| over addition but in some special cases. Moreover, the product has
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| zero divisors and it is defined only on a proper subset of ${\mathbb{R}}^{3}$.
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| %%
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| The construction can be easily generalized to any dimension, but many
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| of its properties are straightforward generalizations of the $3$-dimensional
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| case.
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|
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|
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| \section{Three-dimensional scator algebra}
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|
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| Consider the set of all expressions of the form
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| \[
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| \overset{o}{\varphi}=(f_{0};f_{1},f_{2})\in{\mathbb{R}}^{1+2}\qquad\text{with \ensuremath{f_{i}\in{\mathbb{R}}}},
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| \]
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| with addition defined componentwise. The product of two elements $\overset{o}{\alpha}=(a_{0};a_{1},a_{2}),\overset{o}{\beta}=(b_{0};b_{1},b_{2})$
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| in ${\mathbb{R}}^{1+2}$, say $\overset{o}{\alpha}\cdot\overset{o}{\beta}=\overset{o}{\gamma}=(g_{0};g_{1},g_{2})\in{\mathbb{R}}^{1+2}$
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| is given by \begin{subequations}
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| \begin{eqnarray}
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| g_{0} & = & a_{0}b_{0}+a_{1}b_{1}+a_{2}b_{2}+\frac{a_{1}b_{1}a_{2}b_{2}}{a_{0}b_{0}}=a_{0}b_{0}\Big(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\Big)\Big(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\Big)\qquad\quad\label{eq:product def scalar}\\
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| g_{1} & = & \Big(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\Big)(a_{0}b_{1}+a_{1}b_{0})\label{eq:product def dir1}\\
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| g_{2} & = & \Big(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\Big)(a_{0}b_{2}+a_{2}b_{0}).\label{eq:product def dir2}
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| \end{eqnarray}
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| \end{subequations} In this formulation, the condition for the definition
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| of a product is that
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| \begin{equation}
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| a_{0}b_{0}\neq0\quad\textrm{if}\quad a_{1}b_{1}\neq0\;\text{and}\; a_{2}b_{2}\neq0.\label{eq:condexistencia producto}
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| \end{equation}
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| We call the elements $\overset{o}{\varphi}=(f_{0};f_{1},f_{2})$ \textit{scators}
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| and we will show that, with some restrictions they form a weak hypercomplex
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| system. The scator elements are labeled with an '\textbackslash{}overset\{o\}'.
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| This decoration is omitted when no confusion arises with the scator
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| components.
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| \begin{rem}
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| \label{remark1} To begin with, it is immediate that: \medskip{}
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|
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|
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| \noindent {(1)}: We can identify the elements of the field ${\mathbb{R}}$
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| with the scators of the form $\overset{o}{\alpha}=(a_{0};0,0)$, since
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| addition is componentwise and multiplication of two of these elements,
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| say $\overset{o}{\alpha}$ and $\overset{o}{\beta}=(b_{0};0,0)$,
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| assuming that $a_{0}b_{0}\neq0$, is given by
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| \[
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| \overset{o}{\alpha}\cdot\overset{o}{\beta}=(a_{0};0,0)\cdot(b_{0};0,0)=(a_{0}b_{0};(1+0)(0),(1+0)(0))=(a_{0}b_{0};0,0).
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| \]
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| Sometimes we will use the identification $(a_{0};0,0)=a_{0}$ and
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| call the scator $(a_{0};0,0)$ a \textit{scalar}. \medskip{}
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|
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|
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| \noindent {(2)}: If $\overset{o}{\alpha}=(a_{0};0,0)$ is identified
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| with $a_{0}\in{\mathbb{R}}-\{0\}$ and if $\overset{o}{\beta}=(b_{0};b_{1},b_{2})$
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| is any scator (with $a_{0}b_{0}\neq0$), we have
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| \begin{align*}
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| \overset{o}{\alpha}\cdot\overset{o}{\beta} & =(a_{0};0,0)\cdot(b_{0};b_{1},b_{2})\\
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| & =(a_{0}b_{0};(1)(a_{0}b_{1}+0\cdot b_{0}),(1)(a_{0}b_{2}+0b_{0}))\\
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| & =(a_{0}b_{0};a_{0}b_{1},a_{0}b_{2}).
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| \end{align*}
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| In other words, the product of a scalar with a scator is componentwise.
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| \medskip{}
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|
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|
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| \noindent {(3)}: In particular, for the scalar $(1;0,0)=1\in{\mathbb{R}}$
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| and for any scator $\overset{o}{\varphi}=(f_{0};f_{1},f_{1})$ we
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| have that $1$ is neutral, i.e., $1\cdot\overset{o}{\varphi}=\overset{o}{\varphi}$.
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| \medskip{}
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|
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|
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| \noindent {(4)} The product of a scalar with the sum of two scators
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| is distributive. \medskip{}
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|
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|
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| \noindent {(5)}: The product of the sum of two scalars with a scator
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| is distributive. \medskip{}
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|
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|
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| \noindent {(6)}: Whenever is defined, the product of two scators
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| is commutative. This follows from the symmetry of the definition of
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| product. \medskip{}
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|
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| \end{rem}
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| These remarks mean that the set of scators behaves as an algebra over
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| the real field. However, we have the restriction for the product and,
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| as we shall see, in general the scator product is not distributive,
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| and there are zero divisors. \bigskip{}
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|
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|
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| \noindent \textbf{\label{Conjugate-and-norm.}Conjugate and norm.}
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| The \textit{conjugate} of $\overset{o}{\varphi}=(f_{0};f_{1},f_{2})$
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| is
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| \[
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| \overset{o}{\varphi}^{*}=(f_{0};-f_{1},-f_{2})
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| \]
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| and it is immediate that $(\overset{o}{\varphi}^{*})^{*}=\overset{o}{\varphi}$,
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| $(\overset{o}{\alpha}+\overset{o}{\beta})^{*}=\overset{o}{\alpha}^{*}+\overset{o}{\beta}^{*}$
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| and $(\overset{o}{\alpha}\overset{o}{\beta})^{*}=\overset{o}{\alpha}^{*}\overset{o}{\beta}^{*}$.
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| Moreover, $\overset{o}{\varphi}\in{\mathbb{R}}$ if and only if $\overset{o}{\varphi}^{*}=\overset{o}{\varphi}$.
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|
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| The \textit{norm} or \textit{modulus} squared of $\overset{o}{\varphi}=(f_{0};f_{1},f_{2})$
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| is
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| \begin{equation}
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| \|\overset{o}{\varphi}\|^{2}=\overset{o}{\varphi}\cdot\overset{o}{\varphi}^{*}=f_{0}^{2}\left(1-\frac{f_{1}^{2}}{f_{0}^{2}}\right)\left(1-\frac{f_{2}^{2}}{f_{0}^{2}}\right).\label{eq:sca mag}
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| \end{equation}
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| Thus, the squared norm of a scator is a scalar, $\|\overset{o}{\varphi}\|=\|\overset{o}{\varphi}^{*}\|$
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| and if $\overset{o}{\varphi}=(f_{0};f_{1},f_{2})$ we have that
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| \[
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| f_{0}=\frac{1}{2}\Big(\overset{o}{\varphi}+\overset{o}{\varphi}^{*}\Big).
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| \]
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| Moreover, a direct computation shows that the norm of a product is
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| the product of the norms, provided that zero divisors are excluded.
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| \medskip{}
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|
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|
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| \noindent \textbf{Units.} Given two scators $\overset{o}{\alpha}=(a_{0};a_{1},a_{2}),\overset{o}{\beta}=(b_{0};b_{1},b_{2})$
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| with $a_{0}b_{0}\neq0$, for its product to be $1$ we need that the
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| first component $g_{0}=1$ and the other components $g_{1}=g_{2}=0$:
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| \begin{subequations}
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| \begin{equation}
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| g_{0}=a_{0}b_{0}+a_{1}b_{1}+a_{2}b_{2}+\frac{a_{1}b_{1}a_{2}b_{2}}{a_{0}b_{0}}=a_{0}b_{0}\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)=1\label{eq:product def scalar unit}
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| \end{equation}
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| and the director components of the product are
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| \begin{equation}
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| g_{1}=\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)\left(b_{0}a_{1}+a_{0}b_{1}\right)=0,\label{eq:product def dir1 unit}
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| \end{equation}
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|
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| \begin{equation}
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| g_{2}=\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)\left(b_{0}a_{2}+a_{0}b_{2}\right)=0.\label{eq:product def dir2 unit}
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| \end{equation}
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| \end{subequations} Observe now that if the components of the first
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| factor in \eqref{eq:product def dir1 unit} and \eqref{eq:product def dir2 unit}
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| are zero, i.e., if $\left(1+\frac{a_{j}b_{j}}{a_{0}b_{0}}\right)=0$,
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| then the scalar component is also zero, and thus this possibility
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| is excluded. It follows that the second factors must be zero:
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| \[
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| \left(b_{0}a_{1}+a_{0}b_{1}\right)=0\Rightarrow b_{1}=-\frac{b_{0}a_{1}}{a_{0}}
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| \]
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| and
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| \[
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| \left(b_{0}a_{2}+a_{0}b_{2}\right)=0\Rightarrow b_{2}=-\frac{b_{0}a_{2}}{a_{0}}.
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| \]
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| The first component becomes
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|
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| \[
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| g_{0}=a_{0}b_{0}+a_{1}b_{1}+a_{2}b_{2}+\frac{a_{1}b_{1}a_{2}b_{2}}{a_{0}b_{0}}=a_{0}b_{0}\left(1-\left(\frac{a_{1}}{a_{0}}\right)^{2}\right)\left(1-\left(\frac{a_{2}}{a_{0}}\right)^{2}\right)=1,
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| \]
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| from where it follows that
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| \[
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| b_{0}=\frac{1}{a_{0}\left(1-\left(\frac{a_{1}}{a_{0}}\right)^{2}\right)\left(1-\left(\frac{a_{2}}{a_{0}}\right)^{2}\right)}
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| \]
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| and hence
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| \[
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| b_{1}=-\frac{b_{0}a_{1}}{a_{0}}=-\frac{a_{1}}{a_{0}^{2}\left(1-\left(\frac{a_{1}}{a_{0}}\right)^{2}\right)\left(1-\left(\frac{a_{2}}{a_{0}}\right)^{2}\right)}
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| \]
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| and
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| \[
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| b_{2}=-\frac{b_{0}a_{2}}{a_{0}}=-\frac{a_{2}}{a_{0}^{2}\left(1-\left(\frac{a_{1}}{a_{0}}\right)^{2}\right)\left(1-\left(\frac{a_{2}}{a_{0}}\right)^{2}\right)}.
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| \]
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| Now observe that we must exclude $a_{0}=0$, $\frac{a_{1}}{a_{0}}=\pm1,\frac{a_{2}}{a_{0}}=\pm1$.
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| We have thus shown that $\overset{o}{\beta}=(b_{0};b_{1},b_{2})$
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| is the inverse of $\overset{o}{\alpha}=(a_{0};a_{1},a_{2})$. This
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| can be rewritten as follows. Consider the conjugate $\overset{o}{\alpha}^{*}=(a_{0};-a_{1},-a_{2})$
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| of $\overset{o}{\alpha}$. Then, the product
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| \[
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| \overset{o}{\alpha}\overset{o}{\alpha}^{*}=\big(a_{0}^{2}-a_{1}^{2}-a_{2}^{2}+\frac{a_{1}^{2}a_{2}^{2}}{a_{0}^{2}};0,0\big)
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| \]
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| and so the inverse of $\overset{o}{\alpha}$ is
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| \[
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| \overset{o}{\alpha}^{-1}=\frac{1}{a_{0}^{2}\big(1-\frac{a_{1}^{2}}{a_{0}^{2}}\big)\big(1-\frac{a_{2}^{2}}{a_{0}^{2}}\big)}\overset{o}{\alpha}^{*}
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| \]
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| provided that $\big(1-\frac{a_{1}^{2}}{a_{0}^{2}}\big)\big(1-\frac{a_{2}^{2}}{a_{0}^{2}}\big)\neq0$.
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| Summarizing, the scator $\overset{o}{\alpha}=(a_{0};a_{1},a_{2})$
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| is a unit provided that
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| \begin{equation}
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| a_{0}\neq\pm a_{1},a_{0}\neq\pm a_{2},\quad a_{0}\neq0\quad\text{when}\quad a_{1}a_{2}\neq0.\label{eq:excl no unidades}
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| \end{equation}
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| It follows that $(\overset{o}{\alpha}^{*})^{-1}=(\overset{o}{\alpha}^{-1})^{*}$.
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| \medskip{}
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|
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|
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| \noindent \textbf{Zero divisors.} \label{sec:Divisores-de-cero} Given
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| two non-zero scators ${\overset{o}{\alpha}}=\left(a_{0};a_{1},a_{2}\right)$
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| and ${\overset{o}{\beta}}=\left(b_{0};b_{1},b_{2}\right)$ such that
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| its product ${\overset{o}{\gamma}}={\overset{o}{\alpha}}{\overset{o}{\beta}}=\left(g_{0};g_{1},g_{2}\right)=(0;0,0)$,
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| we want to characterize these zero divisors. From the definition of
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| the scator product \begin{subequations}
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| \begin{equation}
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| g_{0}=a_{0}b_{0}+a_{1}b_{1}+a_{2}b_{2}+\frac{a_{1}b_{1}a_{2}b_{2}}{a_{0}b_{0}}=a_{0}b_{0}\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)=0,\label{eq:product def scalar divcero}
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| \end{equation}
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| where the director components of the product are
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| \begin{equation}
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| g_{1}=\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)\left(b_{0}a_{1}+a_{0}b_{1}\right)=0,\label{eq:product def dir1 divcero}
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| \end{equation}
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|
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| \begin{equation}
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| g_{2}=\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)\left(b_{0}a_{2}+a_{0}b_{2}\right)=0.\label{eq:product def dir2 divcero}
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| \end{equation}
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| \end{subequations} Condition \eqref{eq:product def scalar divcero}
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| is fulfilled if one of the terms between parenthesis is zero: \begin{subequations}
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| \begin{eqnarray}
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| a_{1}b_{1} & = & -a_{0}b_{0}\label{eq:divcero 1}\\
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| a_{2}b_{2} & = & -a_{0}b_{0}\label{eq:divcero 2}
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| \end{eqnarray}
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| \end{subequations} Now, each of the terms between parenthesis is
| |
| also a factor for a $g_{2}$ and $g_{1}$ respectively. Therefore,
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| if both terms are zero, the three components $g_{0},g_{1},g_{2}$
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| are zero. Thus, two factors that satisfy
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| \begin{equation}
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| a_{1}b_{1}=a_{2}b_{2}=-a_{0}b_{0}\label{eq:divcero}
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| \end{equation}
| |
| are zero divisors. There is another case for $g_{0}$ to be zero,
| |
| namely if $a_{0}$ or $b_{0}$ is zero. If $a_{0}=0$ in a non-zero
| |
| scator $\overset{o}{\alpha}$, only one of the director components
| |
| is non-zero for the product to be well-defined, say $a_{0}=0,\: a_{1}\neq0,\: a_{2}=0$.
| |
| In this case, from \eqref{eq:product def dir2 divcero}, the component
| |
| $g_{2}=0$ and the first component of \eqref{eq:product def dir1 divcero},
| |
| must satisfy
| |
| \[
| |
| g_{1}=b_{0}a_{1}=0,\Rightarrow b_{0}=0.
| |
| \]
| |
|
| |
|
| |
| If $b_{0}=0$, the scator ${\overset{o}{\beta}}$ must have only a
| |
| non-zero component, say $b_{1}$. Then the scalar component is non-zero
| |
| $g_{0}=a_{1}b_{1}$. If the non-zero component is $b_{2}$ then the
| |
| components of ${\overset{o}{\gamma}}$ are zero. Summarizing ${\overset{o}{\gamma}}=\left(0;0,0\right)=\left(0;a_{1},0\right)\cdot\left(0;0,b_{2}\right)$.
| |
| Another possibility with these requirements is when ${\overset{o}{\alpha}}=\left(0;0,a_{2}\right)$
| |
| and ${\overset{o}{\beta}}=\left(0;b_{1},0\right)$. These two cases
| |
| are included in \eqref{eq:divcero 1}, \eqref{eq:divcero 2}, because
| |
| if $a_{0}=b_{0}=0$ then $a_{1}b_{1}=0$ y $a_{2}b_{2}=0$ and these
| |
| are non-zero if $a_{1}=b_{2}=0$ or $a_{2}=b_{1}=0$.
| |
|
| |
| So in order to avoid zero divisors we must have \begin{subequations}
| |
| \begin{eqnarray}
| |
| a_{1}b_{1} & \neq & -a_{0}b_{0}\label{eq:no divcero 1}\\
| |
| a_{2}b_{2} & \neq & -a_{0}b_{0}.\label{eq:no divcero 2}
| |
| \end{eqnarray}
| |
| \end{subequations} A particular case of these equations is when
| |
| $b_{1}=-a_{1}$, $a_{0}=b_{0}$, then from \eqref{eq:no divcero 1}
| |
| $a_{1}^{2}\neq a_{0}^{2},$ which is one of the conditions for the
| |
| existence of inverses. The last case for the existence of zero-divisors
| |
| is when one factor $\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)\neq0$
| |
| and the other factor $\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)=0$.
| |
| This last equality implies that $g_{0}=g_{2}=0$ and so, in order
| |
| to have $g_{1}=0$, we must have that $\left(b_{0}a_{1}+a_{0}b_{1}\right)=0\Rightarrow\frac{a_{1}}{a_{0}}=-\frac{b_{1}}{b_{0}}$
| |
| since $\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)=0$, and so
| |
| \[
| |
| \left(1-\frac{a_{1}^{2}}{a_{0}^{2}}\right)=0\Rightarrow a_{1}^{2}=a_{0}^{2}
| |
| \]
| |
| and $b_{1}^{2}=b_{0}^{2}$. Therefore, if $a_{1}^{2}=a_{0}^{2}$
| |
| y $\frac{a_{1}}{a_{0}}=-\frac{b_{1}}{b_{0}}$, then, for any $a_{0},b_{0},a_{2},b_{2}$
| |
| the product is zero. Hence, the condition \eqref{eq:divcero} becomes
| |
| $a_{1}b_{1}=-a_{0}b_{0}$ but we do not have the condition $a_{2}b_{2}=-a_{0}b_{0}$,
| |
| that is $a_{2}b_{2}\neq-a_{0}b_{0}$. However, in this case we have
| |
| that $a_{1}b_{0}=-a_{0}b_{1}$, and dividing
| |
| \[
| |
| \frac{a_{1}b_{1}}{a_{1}b_{0}}=\frac{-a_{0}b_{0}}{-a_{0}b_{1}}\Rightarrow b_{1}^{2}=b_{0}^{2}
| |
| \]
| |
| we obtain
| |
| \[
| |
| {\overset{o}{\gamma}}=\left(0;0,0\right)=\left(a_{0};\pm a_{0},a_{2}\right)\cdot\left(b_{0};\mp b_{0},b_{2}\right)
| |
| \]
| |
| and
| |
| \[
| |
| {\overset{o}{\gamma}}=\left(0;0,0\right)=\left(a_{0};a_{1},\pm a_{0}\right)\cdot\left(b_{0};b_{1},\mp b_{0}\right)
| |
| \]
| |
| and thus the corresponding zero divisor conditions are
| |
| \[
| |
| a_{1}b_{1}=-a_{0}b_{0},\; a_{1}^{2}=a_{0}^{2}
| |
| \]
| |
| and
| |
| \[
| |
| a_{2}b_{2}=-a_{0}b_{0},\; a_{2}^{2}=a_{0}^{2}.
| |
| \]
| |
| Notice that if only units are considered, these last two cases are
| |
| excluded since \eqref{eq:excl no unidades} imposes that the scalar
| |
| components should not be equal to any director component. We have
| |
| thus exhausted all the possibilities for zero divisors in ${\mathbb{R}}^{1+2}$.
| |
|
| |
|
| |
|
| |
| \section{Associativity}
| |
| \begin{lem}
| |
| The product is associative, when divisors of zero are excluded.\end{lem}
| |
| \begin{proof}
| |
| Evaluate ${\zeta}=\left(h_{0};h_{1},h_{2}\right)={\overset{o}{\gamma}}{\overset{o}{\varphi}}=\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}$.
| |
| The scalar component from \eqref{eq:product def scalar} is
| |
| \[
| |
| h_{0}=\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{0}=g_{0}f_{0}\left(1+\frac{g_{1}f_{1}}{g_{0}f_{0}}\right)\left(1+\frac{g_{2}f_{2}}{g_{0}f_{0}}\right),
| |
| \]
| |
| and in turn, evaluation of ${\overset{o}{\gamma}}$ in terms of the
| |
| product ${\overset{o}{\alpha}}{\overset{o}{\beta}}$ gives
| |
|
| |
| \begin{multline*}
| |
| h_{0}=\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{0}=a_{0}b_{0}\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)f_{0}\\
| |
| \left(1+\frac{\left(\frac{a_{1}}{a_{0}}+\frac{b_{1}}{b_{0}}\right)f_{1}}{\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)f_{0}}\right)\left(1+\frac{\left(\frac{a_{2}}{a_{0}}+\frac{b_{2}}{b_{0}}\right)f_{2}}{\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)f_{0}}\right).
| |
| \end{multline*}
| |
| This expression may be rewritten in the nicely symmetrical form
| |
|
| |
| \begin{multline*}
| |
| h_{0}=\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{0}=\\
| |
| a_{0}b_{0}f_{0}\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}+\frac{a_{1}}{a_{0}}\frac{f_{1}}{f_{0}}+\frac{b_{1}}{b_{0}}\frac{f_{1}}{f_{0}}\right)\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}+\frac{a_{2}}{a_{0}}\frac{f_{2}}{f_{0}}+\frac{b_{2}}{b_{0}}\frac{f_{2}}{f_{0}}\right),
| |
| \end{multline*}
| |
| provided that the factors that are being canceled out in numerator
| |
| and denominator are not zero, namely
| |
| \begin{equation}
| |
| \frac{a_{1}b_{1}}{a_{0}b_{0}}\neq-1,\quad\frac{a_{2}b_{2}}{a_{0}b_{0}}\neq-1.\label{eq:not divcero AB}
| |
| \end{equation}
| |
| On the other hand, evaluate ${\zeta}'=\left(h_{0}';h_{1}',h_{2}'\right)={\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)$
| |
| and to ease the notation, set $\tau={\overset{o}{\beta}}{\overset{o}{\varphi}}=(t_{0};t_{1},t_{2})$.
| |
| The scalar component from \eqref{eq:product def scalar} is
| |
| \[
| |
| h_{0}'=\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{0}=a_{0}t_{0}\left(1+\frac{a_{1}t_{1}}{a_{0}t_{0}}\right)\left(1+\frac{a_{2}t_{2}}{a_{0}t_{0}}\right),
| |
| \]
| |
| and in turn, evaluation of ${\tau}$ in terms of the product ${\overset{o}{\beta}}{\overset{o}{\varphi}}$
| |
| gives
| |
|
| |
| \begin{multline*}
| |
| h_{0}'=\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{0}=a_{0}b_{0}f_{0}\left(1+\frac{b_{1}f_{1}}{b_{0}f_{0}}\right)\left(1+\frac{b_{2}f_{2}}{b_{0}f_{0}}\right)\\
| |
| \left(1+\frac{\left(\frac{b_{1}}{b_{0}}+\frac{f_{1}}{f_{0}}\right)a_{1}}{\left(1+\frac{b_{1}f_{1}}{b_{0}f_{0}}\right)a_{0}}\right)\left(1+\frac{\left(\frac{b_{2}}{b_{0}}+\frac{f_{2}}{f_{0}}\right)a_{1}}{\left(1+\frac{b_{2}f_{2}}{b_{0}f_{0}}\right)a_{0}}\right).
| |
| \end{multline*}
| |
| This expression may be rewritten in the symmetrical form
| |
|
| |
| \begin{multline*}
| |
| h_{0}'=\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{0}=\\
| |
| a_{0}b_{0}f_{0}\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}+\frac{a_{1}}{a_{0}}\frac{f_{1}}{f_{0}}+\frac{b_{1}}{b_{0}}\frac{f_{1}}{f_{0}}\right)\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}+\frac{a_{2}}{a_{0}}\frac{f_{2}}{f_{0}}+\frac{b_{2}}{b_{0}}\frac{f_{2}}{f_{0}}\right),
| |
| \end{multline*}
| |
| provided that the terms $\frac{b_{1}f_{1}}{b_{0}f_{0}}$ and $\frac{b_{2}f_{2}}{b_{0}f_{0}}$
| |
| are different from minus one,
| |
| \begin{equation}
| |
| \frac{b_{1}f_{1}}{b_{0}f_{0}}\neq-1,\quad\frac{b_{2}f_{2}}{b_{0}f_{0}}\neq-1.\label{eq:not divcero BF}
| |
| \end{equation}
| |
| Comparison of the expressions for $h_{0}'$ and $h_{0}$, shows that
| |
| they are identical, hence the scalar component of the product is associative
| |
| $\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{0}=\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{0}$.
| |
| Consider now the first director component from \eqref{eq:product def dir1}
| |
| \[
| |
| h_{1}=\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{1}=\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{0}\frac{\left(\frac{g_{1}}{g_{0}}+\frac{f_{1}}{f_{0}}\right)}{\left(1+\frac{g_{1}f_{1}}{g_{0}f_{0}}\right)},
| |
| \]
| |
| evaluation of ${\overset{o}{\gamma}}$ in terms of ${\overset{o}{\alpha}}{\overset{o}{\beta}}$
| |
| components gives
| |
| \[
| |
| \frac{h_{1}}{h_{0}}=\frac{\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{1}}{\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{0}}=\frac{\left(\frac{\left(\frac{a_{1}}{a_{0}}+\frac{b_{1}}{b_{0}}\right)}{\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)}+\frac{f_{1}}{f_{0}}\right)}{\left(1+\frac{\left(\frac{a_{1}}{a_{0}}+\frac{b_{1}}{b_{0}}\right)}{\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)}\frac{f_{1}}{f_{0}}\right)}.
| |
| \]
| |
| This expression may also be put in a symmetrical fashion
| |
| \[
| |
| \frac{h_{1}}{h_{0}}=\frac{\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{1}}{\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{0}}=\frac{\frac{a_{1}}{a_{0}}+\frac{b_{1}}{b_{0}}+\frac{f_{1}}{f_{0}}+\frac{a_{1}b_{1}f_{1}}{a_{0}b_{0}f_{0}}}{1+\frac{a_{1}b_{1}}{a_{0}b_{0}}+\frac{a_{1}f_{1}}{a_{0}f_{0}}+\frac{b_{1}f_{1}}{b_{0}f_{0}}},
| |
| \]
| |
| provided that equations \eqref{eq:not divcero AB} are fulfilled.
| |
| On the other hand, evaluate
| |
| \[
| |
| h_{1}'=\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{1}=\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{0}\frac{\left(\frac{a_{1}}{a_{0}}+\frac{t_{1}}{t_{0}}\right)}{\left(1+\frac{a_{1}t_{1}}{a_{0}t_{0}}\right)},
| |
| \]
| |
| expressing $\tau$ in terms of ${\overset{o}{\beta}}{\overset{o}{\varphi}}$
| |
| components gives
| |
| \[
| |
| \frac{h_{1}'}{h_{0}'}=\frac{\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{1}}{\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{0}}=\frac{\left(\frac{a_{1}}{a_{0}}+\frac{\left(\frac{b_{1}}{b_{0}}+\frac{f_{1}}{f_{0}}\right)}{\left(1+\frac{b_{1}f_{1}}{b_{0}f_{0}}\right)}\right)}{\left(1+\frac{a_{1}}{a_{0}}\frac{\left(\frac{b_{1}}{b_{0}}+\frac{f_{1}}{f_{0}}\right)}{\left(1+\frac{b_{1}f_{1}}{b_{0}f_{0}}\right)}\right)}.
| |
| \]
| |
| This expression may also be put in a symmetrical fashion
| |
| \[
| |
| \frac{h_{1}'}{h_{0}'}=\frac{\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{1}}{\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{0}}=\frac{\frac{a_{1}}{a_{0}}+\frac{b_{1}}{b_{0}}+\frac{f_{1}}{f_{0}}+\frac{a_{1}b_{1}f_{1}}{a_{0}b_{0}f_{0}}}{1+\frac{a_{1}b_{1}}{a_{0}b_{0}}+\frac{a_{1}f_{1}}{a_{0}f_{0}}+\frac{b_{1}f_{1}}{b_{0}f_{0}}},
| |
| \]
| |
| provided that equations \eqref{eq:not divcero BF} are fulfilled.
| |
| Since $h_{0}'=h_{0}$, the first component is also associative $h_{1}'=h_{1}$,
| |
| $\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{1}=\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{1}$.
| |
| An analogous procedure leads to the second component associativity
| |
| $\left[\left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}\right]_{2}=\left[{\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right)\right]_{2}$.
| |
| Since all components are associative, the product is then associative
| |
| \[
| |
| \left({\overset{o}{\alpha}}{\overset{o}{\beta}}\right){\overset{o}{\varphi}}={\overset{o}{\alpha}}\left({\overset{o}{\beta}}{\overset{o}{\varphi}}\right),
| |
| \]
| |
| provided that conditions \ref{eq:not divcero AB} abd \ref{eq:not divcero BF}
| |
| are satisfied. \end{proof}
| |
| \begin{rem}
| |
| The restrictions $\frac{a_{1}b_{1}}{a_{0}b_{0}}\neq-1,\:\frac{a_{2}b_{2}}{a_{0}b_{0}}\neq-1,\:\frac{b_{1}f_{1}}{b_{0}f_{0}}\neq-1,\:\frac{b_{2}f_{2}}{b_{0}f_{0}}\neq-1$
| |
| required for associativity to hold are in fact, the conditions required
| |
| to avoid divisors of zero.
| |
| \end{rem}
| |
|
| |
| \section{Restricted space}
| |
|
| |
| For the product of two scators to be defined we need to restrict the
| |
| corresponding subspace of ${\mathbb{R}}^{1+2}$.
| |
| \begin{lem}
| |
| The restricted space conditions $a_{0},b_{0}\neq0,$ $a_{0}^{2}>a_{1}^{2},a_{2}^{2}$
| |
| and $b_{0}^{2}>b_{1}^{2},b_{2}^{2}$ are sufficient conditions to
| |
| satisfy group properties under the product operation. \end{lem}
| |
| \begin{proof}
| |
| The stronger condition $a_{0},b_{0}\neq0$, encompasses the condition
| |
| required to have a well defined product \eqref{eq:condexistencia producto}.
| |
| Unit and inverse existance are assured since the inequality of scalar
| |
| components larger than director or versor components prevents the
| |
| equalities $a_{0}^{2}=a_{1}^{2},\: a_{0}^{2}=a_{2}^{2}$. Divisors
| |
| of zero are avoided because the restricted space conditions impose
| |
| $\left(\frac{a_{1}}{a_{0}}\right)^{2}<1$ and $\left(\frac{b_{0}}{b_{1}}\right)^{2}>1$,
| |
| hence $\frac{a_{1}}{a_{0}}\neq\pm\frac{b_{0}}{b_{1}}$ and similarly
| |
| for the second components. Therefore conditions \eqref{eq:not divcero AB}
| |
| and \eqref{eq:not divcero BF} are fulfilled and associativity holds.
| |
| To prove closure under the restricted space domain, recall that $1>\frac{a_{j}}{a_{0}}$,
| |
| thus $1-\frac{a_{j}}{a_{0}}>0$, similarly $1-\frac{b_{j}}{b_{0}}>0$.
| |
| The product of these two terms gives the inequality
| |
| \[
| |
| \left(1-\frac{a_{j}}{a_{0}}\right)\left(1-\frac{b_{j}}{b_{0}}\right)>0,\: j=1,2.
| |
| \]
| |
| The expansion of this product is $1-\frac{a_{j}}{a_{0}}-\frac{b_{j}}{b_{0}}+\frac{a_{j}}{a_{0}}\frac{b_{j}}{b_{0}}>0$,
| |
| so that $1+\frac{a_{j}}{a_{0}}\frac{b_{j}}{b_{0}}>\frac{a_{j}}{a_{0}}+\frac{b_{j}}{b_{0}}$.
| |
| It is possible to divide by $1+\frac{a_{j}}{a_{0}}\frac{b_{j}}{b_{0}}$
| |
| since this term is not zero,
| |
| \[
| |
| 1>\frac{\frac{a_{j}}{a_{0}}+\frac{b_{j}}{b_{0}}}{1+\frac{a_{j}}{a_{0}}\frac{b_{j}}{b_{0}}}.
| |
| \]
| |
| Comparison with equations \eqref{eq:product def scalar}, \eqref{eq:product def dir1}
| |
| and \eqref{eq:product def dir2} shows that the RHS of the inequality
| |
| is the quotient $\frac{g_{j}}{g_{0}}$ of the product of two scators,
| |
| thus $g_{0}>g_{j}$. The domain and the codomain of the restricted
| |
| space are then equal.\end{proof}
| |
| \begin{rem}
| |
| There is a different restricted space condition, when the director
| |
| components terms are larger than the scalar term $a_{0},b_{0}\neq0,$
| |
| $a_{0}^{2}<a_{1}^{2},a_{2}^{2}$ and $b_{0}^{2}<b_{1}^{2},b_{2}^{2}$,
| |
| where the scator product also satisfies group properties.
| |
| \end{rem}
| |
|
| |
|
| |
|
| |
| \section{Scator in terms of a basis}
| |
|
| |
| The scators $1,\hat{\mathbf{e}}_{1}=(0;1,0),\hat{\mathbf{e}}_{2}=(0;0,1)$
| |
| form a basis for ${\mathbb{R}}^{1+2}$ and so any scator may be written
| |
| as
| |
| \[
| |
| {\overset{o}{\alpha}}=\left(a_{0};a_{1},a_{2}\right)=a_{0}+a_{1}\hat{\mathbf{e}}_{1}+a_{2}\hat{\mathbf{e}}_{2}.
| |
| \]
| |
| The first component of the first element is necessarily a scalar
| |
| as we have already proved in item (1) of remark \ref{remark1}. The
| |
| product of two scators ${\overset{o}{\alpha}}{\overset{o}{\beta}}$
| |
| according to the definition \eqref{eq:product def scalar}-\eqref{eq:product def dir2}
| |
| is
| |
| \begin{multline}
| |
| {\overset{o}{\alpha}}{\overset{o}{\beta}}=\left(a_{0}b_{0}+a_{1}b_{1}+a_{2}b_{2}+\frac{a_{1}b_{1}a_{2}b_{2}}{a_{0}b_{0}}\right)\\
| |
| +\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)\left(b_{0}a_{1}+a_{0}b_{1}\right)\hat{\mathbf{e}}_{1}+\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)\left(b_{0}a_{2}+a_{0}b_{2}\right)\hat{\mathbf{e}}_{2}\label{eq:product versors}
| |
| \end{multline}
| |
| This operation is not bilinear and therefore cannot be written in
| |
| terms of the product of its addends. To wit, if the product were distributed
| |
| over the scator components the following nine terms would be obtained
| |
| \begin{multline*}
| |
| {\overset{o}{\alpha}}{\overset{o}{\beta}}\overset{?}{=}a_{0}b_{0}+a_{0}b_{1}\hat{\mathbf{e}}_{1}+a_{0}b_{2}\hat{\mathbf{e}}_{2}+b_{0}a_{1}\hat{\mathbf{e}}_{1}+b_{0}a_{2}\hat{\mathbf{e}}_{2}\\
| |
| +b_{1}a_{1}\hat{\mathbf{e}}_{1}\hat{\mathbf{e}}_{1}+b_{2}a_{1}\hat{\mathbf{e}}_{1}\hat{\mathbf{e}}_{2}+b_{1}a_{2}\hat{\mathbf{e}}_{2}\hat{\mathbf{e}}_{1}+b_{2}a_{2}\hat{\mathbf{e}}_{2}\hat{\mathbf{e}}_{2}.
| |
| \end{multline*}
| |
| However, each of the director products $\hat{\mathbf{e}}_{1}\hat{\mathbf{e}}_{1},\hat{\mathbf{e}}_{1}\hat{\mathbf{e}}_{2},\hat{\mathbf{e}}_{2}\hat{\mathbf{e}}_{1},\hat{\mathbf{e}}_{2}\hat{\mathbf{e}}_{2}$
| |
| cannot be rewritten in the general form $x+y\hat{\mathbf{e}}_{1}+z\hat{\mathbf{e}}_{2}$
| |
| with $x,y,z$ independent of the scator coefficients in order to reproduce
| |
| the required product terms in (\ref{eq:product versors}). \begin{lem}
| |
| Hyperbolic $1+1$ dimensional scators, that is with only one non-vanishing
| |
| director component either $\hat{\mathbf{e}}_{1}$ or $\hat{\mathbf{e}}_{2}$,
| |
| are identical to double numbers, as defined in \cite{KantorSolodovnikov:1989}.
| |
| \end{lem} \begin{proof} Consider the particular case of the product
| |
| of two scators with only one and the same non zero director component,
| |
| say $\hat{\mathbf{e}}_{2}$ (but could equally be $\hat{\mathbf{e}}_{1}$),
| |
| from the scator product definition \eqref{eq:product def scalar}-\eqref{eq:product def dir2}
| |
| \[
| |
| {\overset{o}{\alpha}}{\overset{o}{\beta}}=\left(a_{0}b_{0}+a_{2}b_{2};0,b_{0}a_{2}+a_{0}b_{2}\right)=a_{0}b_{0}+a_{2}b_{2}+\left(b_{0}a_{2}+a_{0}b_{2}\right)\hat{\mathbf{e}}_{2}
| |
| \]
| |
| The product thus becomes identical to the product of hyperbolic complex
| |
| or double numbers. The product can then be distributed over the scator
| |
| components
| |
| \[
| |
| {\overset{o}{\alpha}}{\overset{o}{\beta}}=\left(a_{0}+a_{2}\hat{\mathbf{e}}_{2}\right)\left(b_{0}+b_{2}\hat{\mathbf{e}}_{2}\right)=a_{0}b_{0}+a_{2}b_{2}+\left(b_{0}a_{2}+a_{0}b_{2}\right)\hat{\mathbf{e}}_{2}
| |
| \]
| |
| provided that $\hat{\mathbf{e}}_{2}\hat{\mathbf{e}}_{2}=1$. The
| |
| addition operation is the same component wise for scators and double
| |
| numbers. 1+1 hyperbolic scators with either $\hat{\mathbf{e}}_{1}$
| |
| or $\hat{\mathbf{e}}_{2}$ thus form a commutative ring identical
| |
| to double $\mathbb{H}_{2}$ numbers.\end{proof} \begin{rem} The
| |
| equality $\hat{\mathbf{e}}_{2}\hat{\mathbf{e}}_{2}=1$ is consistent
| |
| with the scator product definition (\ref{eq:product def scalar}),
| |
| since the product $\left(0;0,1\right)\left(0;0,1\right)$ is equal
| |
| to $\left(1;0,0\right)$. Care should be taken to make the null director
| |
| component zero before the scalar component is taken to the zero limit.
| |
| \end{rem} On the other hand, if we consider the particular case of
| |
| the product of two scators with different non-vanishing vector component
| |
| from the scator product definition
| |
| \[
| |
| {\overset{o}{\alpha}}{\overset{o}{\beta}}=\left(a_{0};a_{1},0\right)\left(b_{0};0,b_{2}\right)=\left(a_{0}b_{0};b_{0}a_{1},a_{0}b_{2}\right).
| |
| \]
| |
| In terms of the basis $1,\hat{\mathbf{e}}_{1},\hat{\mathbf{e}}_{2}$,
| |
| this operation is
| |
| \[
| |
| {\overset{o}{\alpha}}{\overset{o}{\beta}}=\left(a_{0}+a_{1}\hat{\mathbf{e}}_{1}\right)\left(b_{0}+b_{2}\hat{\mathbf{e}}_{2}\right)=a_{0}b_{0}+a_{1}b_{0}\hat{\mathbf{e}}_{1}+a_{0}b_{2}\hat{\mathbf{e}}_{2}
| |
| \]
| |
|
| |
|
| |
| The product can then be distributed over the scator components provided
| |
| that $\hat{\mathbf{e}}_{1}\hat{\mathbf{e}}_{2}=0$. This last equality
| |
| is consistent with the scator product definition, namely the product
| |
| $\left(0;1,0\right)\left(0;0,1\right)$ is equal to $\left(0;0,0\right)$.
| |
| Then we can construct table \ref{tab:product-of-single} with the
| |
| product of scators with a single non-vanishing component. Notice that
| |
| this table does not contain all the information needed to define the
| |
| product of two arbitrary scators with two non-vanishing components.
| |
|
| |
| \begin{table}
| |
| \noindent \begin{centering}
| |
| \begin{tabular}{|c|c|c|c|}
| |
| \hline
| |
| & 1 & $\hat{\mathbf{e}}_{1}$ & $\hat{\mathbf{e}}_{2}$\tabularnewline
| |
| \hline
| |
| \hline
| |
| 1 & 1 & $\hat{\mathbf{e}}_{1}$ & $\hat{\mathbf{e}}_{2}$\tabularnewline
| |
| \hline
| |
| $\hat{\mathbf{e}}_{1}$ & $\hat{\mathbf{e}}_{1}$ & 1 & 0\tabularnewline
| |
| \hline
| |
| $\hat{\mathbf{e}}_{2}$ & $\hat{\mathbf{e}}_{2}$ & 0 & 1\tabularnewline
| |
| \hline
| |
| \end{tabular}
| |
| \par\end{centering}
| |
|
| |
| \caption{\label{tab:product-of-single} Product of single director elements.
| |
| This table should be read with care since it does not contain all
| |
| the information needed to define the product of two arbitrary scators. }
| |
| \end{table}
| |
|
| |
|
| |
| Recall that the characteristic matrix of a hypercomplex algebra is
| |
| generated with the bi-product of all the components in the director
| |
| basis. The multiplication table \ref{tab:product-of-single} describes
| |
| the product of 1+2 hyperbolic scators with a single non-vanishing
| |
| component. This is as close as we can get to a characteristic matrix.
| |
| However, it is of limited use because the product of two 1+2 hyperbolic
| |
| scators cannot be obtained from this matrix as we shall presently
| |
| show due to the lack of distributivity.
| |
|
| |
| \begin{lem} The product of two arbitrary $1+2$ hyperbolic scators
| |
| cannot be represented by a matrix-matrix product. \end{lem} \begin{proof}
| |
| Suppose that the product of two $1+2$ hyperbolic scators can be represented
| |
| by the product of two matrices $\mathcal{M}_{G}$ and $\mathcal{M}_{F}$,
| |
| that is
| |
| \[
| |
| \mathcal{M}_{GF}=\mathcal{M}_{G}\mathcal{M}_{F}.
| |
| \]
| |
| Hyperbolic scator addition can be represented by the sum of two matrices
| |
| since the sum is evaluated by the sum of each component. Let the matrix
| |
| $\mathcal{M}_{G}$ be the sum of two scators be represented by the
| |
| sum of the matrices $\mathcal{M}_{A}$ and $\mathcal{M}_{B}$, then
| |
| \[
| |
| \mathcal{M}_{GF}=\left(\mathcal{M}_{A}+\mathcal{M}_{B}\right)\mathcal{M}_{F}.
| |
| \]
| |
| However, matrix product is distributive over addition, thus
| |
| \[
| |
| \mathcal{M}_{GF}=\mathcal{M}_{A}\mathcal{M}_{F}+\mathcal{M}_{B}\mathcal{M}_{F}.
| |
| \]
| |
| But we have already proved that the scator product is not distributive
| |
| over addition. Therefore, the scator product cannot be isomorphic
| |
| to matrix multiplication. \end{proof} The concept of modulus of a
| |
| complex number can be extended to hypercomplex numbers by taking the
| |
| $n$-th root of the absolute value of the characteristic determinant
| |
| \cite[ch.2.1.6]{catoni2008}. It is not possible to extend this procedure
| |
| to this case because the characteristic matrix does not contain all
| |
| the information involved in the product of two arbitrary scators.
| |
|
| |
|
| |
| \section{Lack of Distributivity}
| |
|
| |
| The difference between the product $\left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}$
| |
| and the sum of the products ${\overset{o}{\alpha}}{\overset{o}{\gamma}}+{\overset{o}{\beta}}{\overset{o}{\gamma}}$
| |
| is a measure of the lack of distributivity. The product $\left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}$
| |
| is
| |
| \begin{multline*}
| |
| \left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}=\left(a_{0}+b_{0}\right)g_{0}\left(1+\frac{\left(a_{1}+b_{1}\right)g_{1}}{\left(a_{0}+b_{0}\right)g_{0}}\right)\left(1+\frac{\left(a_{2}+b_{2}\right)g_{2}}{\left(a_{0}+b_{0}\right)g_{0}}\right)\\
| |
| \left(1;\frac{\left(\frac{\left(a_{1}+b_{1}\right)}{\left(a_{0}+b_{0}\right)}+\frac{g_{1}}{g_{0}}\right)}{\left(1+\frac{\left(a_{1}+b_{1}\right)g_{1}}{\left(a_{0}+b_{0}\right)g_{0}}\right)},\frac{\left(\frac{\left(a_{2}+b_{2}\right)}{\left(a_{0}+b_{0}\right)}+\frac{g_{2}}{g_{0}}\right)}{\left(1+\frac{\left(a_{2}+b_{2}\right)g_{2}}{\left(a_{0}+b_{0}\right)g_{0}}\right)}\right),
| |
| \end{multline*}
| |
| while the product ${\overset{o}{\alpha}}{\overset{o}{\gamma}}$ is
| |
| \[
| |
| {\overset{o}{\alpha}}{\overset{o}{\gamma}}=a_{0}g_{0}\left(1+\frac{a_{1}g_{1}}{a_{0}g_{0}}\right)\left(1+\frac{a_{2}g_{2}}{a_{0}g_{0}}\right)\left(1;\frac{\left(\frac{a_{1}}{a_{0}}+\frac{g_{1}}{g_{0}}\right)}{\left(1+\frac{a_{1}g_{1}}{a_{0}g_{0}}\right)},\frac{\left(\frac{a_{2}}{a_{0}}+\frac{g_{2}}{g_{0}}\right)}{\left(1+\frac{a_{2}g_{2}}{a_{0}g_{0}}\right)}\right),
| |
| \]
| |
| and the product ${\overset{o}{\beta}}{\overset{o}{\gamma}}$ is
| |
| \[
| |
| {\overset{o}{\beta}}{\overset{o}{\gamma}}=b_{0}g_{0}\left(1+\frac{b_{1}g_{1}}{b_{0}g_{0}}\right)\left(1+\frac{b_{2}g_{2}}{b_{0}g_{0}}\right)\left(1;\frac{\left(\frac{b_{1}}{b_{0}}+\frac{g_{1}}{g_{0}}\right)}{\left(1+\frac{b_{1}g_{1}}{b_{0}g_{0}}\right)},\frac{\left(\frac{b_{2}}{b_{0}}+\frac{g_{2}}{g_{0}}\right)}{\left(1+\frac{b_{2}g_{2}}{b_{0}g_{0}}\right)}\right).
| |
| \]
| |
|
| |
|
| |
|
| |
| \subsection{Scalar part}
| |
|
| |
| The difference between the product with the sum and the sum of the
| |
| products $\left[\left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}-\left({\overset{o}{\alpha}}{\overset{o}{\gamma}}+{\overset{o}{\beta}}{\overset{o}{\gamma}}\right)\right]_{0}$
| |
| for the scalar part is
| |
| \begin{align*}
| |
| & \left(\left(a_{0}+b_{0}\right)g_{0}+\left(a_{1}+b_{1}\right)g_{1}+\left(a_{2}+b_{2}\right)g_{2}+\frac{\left(a_{1}+b_{1}\right)g_{1}\left(a_{2}+b_{2}\right)g_{2}}{\left(a_{0}+b_{0}\right)g_{0}}\right)\\
| |
| & -\left(a_{0}g_{0}+a_{1}g_{1}+a_{2}g_{2}+\frac{a_{1}g_{1}a_{2}g_{2}}{a_{0}g_{0}}\right)\\
| |
| & -\left(b_{0}g_{0}+b_{1}g_{1}+b_{2}g_{2}+\frac{b_{1}g_{1}b_{2}g_{2}}{b_{0}g_{0}}\right).
| |
| \end{align*}
| |
| This expression simplifies to
| |
| \[
| |
| \left(\frac{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right)}{\left(a_{0}+b_{0}\right)}-\frac{a_{1}a_{2}}{a_{0}}-\frac{b_{1}b_{2}}{b_{0}}\right)\frac{g_{1}g_{2}}{g_{0}}.
| |
| \]
| |
| The term within brackets can be factored as
| |
| \[
| |
| \left(\frac{\left(b_{0}a_{1}-a_{0}b_{1}\right)\left(a_{0}b_{2}-b_{0}a_{2}\right)}{\left(a_{0}+b_{0}\right)a_{0}b_{0}}\right)\frac{g_{1}g_{2}}{g_{0}}.
| |
| \]
| |
|
| |
|
| |
|
| |
| \subsection{Director component 1}
| |
|
| |
| The product of the sum for the first component is
| |
| \[
| |
| \left[\left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}\right]_{1}=\left(a_{0}+b_{0}\right)g_{0}\left(1+\frac{\left(a_{2}+b_{2}\right)g_{2}}{\left(a_{0}+b_{0}\right)g_{0}}\right)\left(\frac{\left(a_{1}+b_{1}\right)}{\left(a_{0}+b_{0}\right)}+\frac{g_{1}}{g_{0}}\right)
| |
| \]
| |
| that may be expanded as
| |
| \begin{multline*}
| |
| \left[\left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}\right]_{1}=\\
| |
| g_{0}\left(a_{1}+b_{1}\right)+\left(a_{0}+b_{0}\right)g_{1}+\frac{\left(a_{2}+b_{2}\right)g_{1}g_{2}}{g_{0}}+\frac{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right)g_{2}}{\left(a_{0}+b_{0}\right)}.
| |
| \end{multline*}
| |
| While the product $\left[{\overset{o}{\alpha}}{\overset{o}{\gamma}}\right]_{1}$
| |
| for the first component is
| |
| \begin{eqnarray*}
| |
| \left[{\overset{o}{\alpha}}{\overset{o}{\gamma}}\right]_{1} & = & a_{0}g_{0}\left(1+\frac{a_{2}g_{2}}{a_{0}g_{0}}\right)\left(\frac{a_{1}}{a_{0}}+\frac{g_{1}}{g_{0}}\right)\\
| |
| & = & a_{1}g_{0}+a_{0}g_{1}+\frac{a_{1}a_{2}g_{2}}{a_{0}}+\frac{g_{1}g_{2}a_{2}}{g_{0}}
| |
| \end{eqnarray*}
| |
| and a similar result holds for $\left[{\overset{o}{\beta}}{\overset{o}{\gamma}}\right]_{1}$
| |
| with the substitution $A\rightarrow B$. The difference for the first
| |
| component $\left[\left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}-\left({\overset{o}{\alpha}}{\overset{o}{\gamma}}+{\overset{o}{\beta}}{\overset{o}{\gamma}}\right)\right]_{1}$
| |
| is then
| |
| \begin{eqnarray*}
| |
| & & \left(g_{0}\left(a_{1}+b_{1}\right)+\left(a_{0}+b_{0}\right)g_{1}+\frac{\left(a_{2}+b_{2}\right)g_{1}g_{2}}{g_{0}}+\frac{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right)g_{2}}{\left(a_{0}+b_{0}\right)}\right)\\
| |
| & & -\left(a_{1}g_{0}+a_{0}g_{1}+\frac{a_{1}a_{2}g_{2}}{a_{0}}+\frac{g_{1}g_{2}a_{2}}{g_{0}}\right)\\
| |
| & & -\left(b_{1}g_{0}+b_{0}g_{1}+\frac{b_{1}b_{2}g_{2}}{b_{0}}+\frac{g_{1}g_{2}b_{2}}{g_{0}}\right).
| |
| \end{eqnarray*}
| |
| This expression simplifies to
| |
| \[
| |
| \left(\frac{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right)}{\left(a_{0}+b_{0}\right)}-\frac{a_{1}a_{2}}{a_{0}}-\frac{b_{1}b_{2}}{b_{0}}\right)g_{2}
| |
| \]
| |
| that may be factored as
| |
| \[
| |
| \left(\frac{\left(b_{0}a_{1}-a_{0}b_{1}\right)\left(a_{0}b_{2}-b_{0}a_{2}\right)}{\left(a_{0}+b_{0}\right)a_{0}b_{0}}\right)g_{2}.
| |
| \]
| |
|
| |
|
| |
|
| |
| \subsection{Director component 2}
| |
|
| |
| The difference $\left[\left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}-\left({\overset{o}{\alpha}}{\overset{o}{\gamma}}+{\overset{o}{\beta}}{\overset{o}{\gamma}}\right)\right]_{2}$
| |
| for the second component follows a similar procedure that gives
| |
|
| |
| \[
| |
| \left(\frac{\left(b_{0}a_{1}-a_{0}b_{1}\right)\left(a_{0}b_{2}-b_{0}a_{2}\right)}{\left(a_{0}+b_{0}\right)a_{0}b_{0}}\right)g_{1}.
| |
| \]
| |
|
| |
| \begin{rem}
| |
| Given scators ${\overset{o}{\alpha}},{\overset{o}{\beta}},{\overset{o}{\gamma}}$
| |
| with 1+2 components the distributivity difference is
| |
|
| |
| \begin{multline}
| |
| \left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}-\left[\left({\overset{o}{\alpha}}{\overset{o}{\gamma}}\right)+\left({\overset{o}{\beta}}{\overset{o}{\gamma}}\right)\right]=\\
| |
| \frac{\left(b_{0}a_{1}-a_{0}b_{1}\right)\left(a_{0}b_{2}-b_{0}a_{2}\right)}{\left(a_{0}+b_{0}\right)a_{0}b_{0}}\left(\frac{g_{1}g_{2}}{g_{0}};g_{2},g_{1}\right)\label{eq: distre1+2}
| |
| \end{multline}
| |
|
| |
|
| |
| Let us evaluate some particular cases:
| |
|
| |
| (1) If the scators ${\overset{o}{\alpha}}$ and ${\overset{o}{\beta}}$
| |
| have only one and the same non-vanishing director component, say the
| |
| second component, then the above expression is zero and the product
| |
| is distributive over addition.
| |
|
| |
| (2) If the scators ${\overset{o}{\alpha}}$ and ${\overset{o}{\beta}}$
| |
| have only one director component but not the same one, the above expression
| |
| is different from zero, for example if $a_{2}=b_{1}=0$,
| |
| \[
| |
| \left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}-\left[\left({\overset{o}{\alpha}}{\overset{o}{\gamma}}\right)+\left({\overset{o}{\beta}}{\overset{o}{\gamma}}\right)\right]=\frac{a_{1}b_{2}}{\left(a_{0}+b_{0}\right)}\left(\frac{g_{1}g_{2}}{g_{0}};g_{2},g_{1}\right)
| |
| \]
| |
|
| |
|
| |
| (3) If the scator ${\overset{o}{\gamma}}=\left\{ g_{0},g_{1},0\right\} $
| |
| has only one component, then
| |
| \[
| |
| \left({\overset{o}{\alpha}}+{\overset{o}{\beta}}\right){\overset{o}{\gamma}}-\left[\left({\overset{o}{\alpha}}{\overset{o}{\gamma}}\right)+\left({\overset{o}{\beta}}{\overset{o}{\gamma}}\right)\right]=\frac{\left(b_{0}a_{1}-a_{0}b_{1}\right)\left(a_{0}b_{2}-b_{0}a_{2}\right)}{\left(a_{0}+b_{0}\right)a_{0}b_{0}}\left(0;0,g_{1}\right).
| |
| \]
| |
|
| |
| \end{rem}
| |
|
| |
| \section{Composition of velocities in a deformed Lorentz metric}
| |
|
| |
| We now examine one of the possible applications of real or hyperbolic
| |
| scator algebra. In the canonical description of special relativity,
| |
| the Lorentz transformations establish the operations required to transform
| |
| time-space between inertial frames. The separation between events
| |
| is chosen as the square root of a quadratic form. The signature of
| |
| this form together with a Minkowskian system of coordinates completes
| |
| the framework \cite{Synge72}. Einstein's theorem of addition of velocities
| |
| is a consequence of this approach. Amendments to the Lorentz invariant
| |
| have been proposed on different grounds, in particular regarding a
| |
| fundamental length at the Planck scale \cite{AmelinoCamelia02,KowalskiGlikman04}
| |
| and deformed metrics depending on the energy and nature of interactions
| |
| \cite{Cardone2007}. The composition of velocities in special relativity
| |
| ought to be consistent with the two fundamental postulates, the constancy
| |
| of the speed of light and the equivalence of all observers in free
| |
| inertial motion.
| |
|
| |
| There is liberty regarding the choice of mathematical structure selected
| |
| in order to model physical phenomena. Poincaré expressed this idea,
| |
| according to Carnap, in the following statement: ``\ldots{} the
| |
| physicist is free to ascribe to physical space any one of the mathematically
| |
| possible geometrical structures, provided he makes suitable adjustments
| |
| in the laws of mechanics and optics and consequently in the rules
| |
| for measuring length.'' \cite{Reichenbach58}. In accordance with
| |
| this view, it should be possible to produce a different set of rules
| |
| to perform an operation, say the product. These rules cannot be arbitrarily
| |
| set, but have to fulfill the constraints of the geometrical structure
| |
| according to the phenomena that are being modeled or described. The
| |
| procedure for measuring magnitudes should also be altered, that is,
| |
| the metric modified. Physical meaning can be ascribed to this operation,
| |
| in this particular proposal, the composition of velocities between
| |
| inertial frames. Let us follow these guidelines. Allow for \eqref{eq:product def scalar}-\eqref{eq:product def dir2}
| |
| to be the new set of product rules. Consider the quotient of the scator
| |
| director components $a_{1}$ and $a_{2}$ over the scalar component
| |
| $a_{0}$ represent the velocity components (divided by the real constant
| |
| c) in the directions 1 and 2, that is $\beta_{a1}\equiv\frac{v_{a1}}{c},\beta_{a2}\equiv\frac{v_{a2}}{c}$.
| |
| The velocity scator is then $\frac{\overset{o}{\alpha}}{a_{0}}=\left(1;\beta_{a1},\beta_{a2}\right)$.
| |
| Similarly, the velocity scators $\overset{o}{\beta}$ and $\overset{o}{\gamma}$
| |
| are $\frac{\overset{o}{\beta}}{b_{0}}=\left(1;\beta_{b1},\beta_{22}\right)$
| |
| and $\frac{\overset{o}{\gamma}}{g_{0}}=\left(1;\beta_{g1},\beta_{g2}\right)$
| |
| respectively. Let the composition of velocities be represented by
| |
| the product operation $\overset{o}{\gamma}=\overset{o}{\alpha}\overset{o}{\beta}=\left(g_{0};g_{1},g_{2}\right)$.
| |
| The velocity component in direction 1 is
| |
| \[
| |
| \beta_{g1}=\frac{g_{1}}{g_{0}}=\frac{\left(\frac{a_{1}}{a_{0}}+\frac{b_{1}}{b_{0}}\right)}{\left(1+\frac{a_{1}b_{1}}{a_{0}b_{0}}\right)}=\frac{\left(\beta_{a1}+\beta_{b1}\right)}{\left(1+\beta_{a1}\beta_{b1}\right)}.
| |
| \]
| |
| The composition of velocities in direction 2 is
| |
| \[
| |
| \beta_{g2}=\frac{g_{2}}{g_{0}}=\frac{\left(\frac{a_{2}}{a_{0}}+\frac{b_{2}}{b_{0}}\right)}{\left(1+\frac{a_{2}b_{2}}{a_{0}b_{0}}\right)}=\frac{\left(\beta_{a2}+\beta_{b2}\right)}{\left(1+\beta_{a2}\beta_{b2}\right)}.
| |
| \]
| |
| From these results, we have recently established the generalization
| |
| for the composition of velocities to three dimensions \cite{fernandez-guasti2011_2}.
| |
| The constraints imposed on the mathematical structure are that the
| |
| product operation forms a commutative group. These constraints are
| |
| established on the following grounds:
| |
| \begin{itemize}
| |
| \item Closure - the composition of velocities should be an admissible velocity.
| |
| \item Neutral - a relative velocity frame should exist that does not alter
| |
| the velocity of the object.
| |
| \item Inverse - there should exist a frame for any event where the object
| |
| is at rest.
| |
| \item Associativity - the composition of velocities should fulfill the reciprocity
| |
| principle \cite{Berzi1968}.
| |
| \item Commutativity - the composition of velocities should fulfill the reciprocity
| |
| principle.
| |
| \end{itemize}
| |
| \noindent In one dimension, the group-theoretic differentiable functions
| |
| are isomorphic to the sum of the reals \cite{zaldivar2011}. In higher
| |
| dimensions \ldots{} Let the definition of magnitude be given by \eqref{eq:sca mag}.
| |
| This magnitude definition is identical to the Lorentz metric $\sqrt{c^{2}t^{2}-x^{2}-y^{2}-z^{2}}$
| |
| in 1+1 dimensions (one spatial dimension). It also approaches the
| |
| Lorentz metric in the paraxial regime. Departures arise in two or
| |
| more spatial dimensions, that is, when the Thomas precession comes
| |
| into play \cite{fernandez-guasti2011_2}. The equation for constant
| |
| metric $m$ is given by $\sqrt{c^{2}t^{2}-x^{2}-y^{2}+\frac{x^{2}y^{2}}{c^{2}t^{2}}}=m$;
| |
| In polynomial form with $c=1$,
| |
| \[
| |
| t^{4}-\left(m^{2}+x^{2}+y^{2}\right)t^{2}+x^{2}y^{2}=0
| |
| \]
| |
| For a light-like event, the constant $m$ is zero. The light curves
| |
| for constant time $t$ are squares with side equal to $2t$ \cite{Fernandezguasti92}.
| |
| The light events surface is composed by two tetragonal pyramids joined
| |
| by their vertices at the origin. A plot of the light pyramid as well
| |
| as the light cone in Minkowski space is shown in figure \ref{fig:cono mink-1}.
| |
| \begin{figure}[ht]
| |
| \begin{centering}
| |
| \includegraphics[scale=0.6]{scator-metric-1-sxy-mink-cono0} \includegraphics[scale=0.65]{scator-metric-1-f012-1}
| |
| \par\end{centering}
| |
|
| |
| \caption{\label{fig:cono mink-1}Minkowski's light cone (left) and light pyramid
| |
| (right) obtained with the scator metric. The vertical axis represents
| |
| the time axis and the two orthogonal axes represent two spatial dimensions.
| |
| The velocity at any point of the pyramid surface is always c as may
| |
| be seen from the scator metric definition. }
| |
| \end{figure}
| |
|
| |
|
| |
| The restricted space condition establishes the function domain, which,
| |
| in the special relativity terminology, corresponds to the admissible
| |
| velocity. For director components larger than the scalar components,
| |
| spatial-like events are described. Hypothetical particles within this
| |
| realm have been named tachyons. For director components smaller than
| |
| the scalar components, time-like events are described. These are the
| |
| events within the light pyramid accessible to an observer located
| |
| at the vertex of the dipyramid. The proposed framework is in accordance
| |
| with the constancy of the speed of light limit regardless of the motion
| |
| of the source. An important feature of the present proposal is that
| |
| the composition of velocities forms a commutative group. The velocity
| |
| reciprocity principle is then immediately fulfilled.
| |
|
| |
|
| |
| \section{Conclusions}
| |
|
| |
| We have presented an algebra structured in $\mathbb{R}^{3}$ with
| |
| componetwise addition and a novel product operation that has zero
| |
| divisors. If these divisors are excluded, both the addition and product
| |
| operations satisfy commutative group properties. However, the product
| |
| does not distribute over addition. Note that in the two dimensonal
| |
| case, distributivity holds and the algebra is iso to hyperbolic complex
| |
| numbers.
| |
|
| |
| The first scator component comports like a real scalar whereas the
| |
| two remaining components are symmetrical and exhibit a behaviour similar
| |
| to quantities with direction. The construction can be generalized
| |
| to any dimension. \foreignlanguage{spanish}{The product of two scators
| |
| $\overset{o}{\alpha}=\left(a_{0};a_{1},a_{2},\ldots,a_{n}\right)$
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| y $\overset{o}{\beta}=\left(b_{0};b_{1},b_{2},\ldots,b_{n}\right)$
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| with arbitrary dimension $n$ is then defined by $\overset{o}{\gamma}=\overset{o}{\alpha}\overset{o}{\beta}=\left(g_{0};g_{1},g_{2},\ldots,g_{n}\right)$,
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| where the scalar component of the product is
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| \begin{equation}
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| g_{0}=a_{0}b_{0}\prod_{k=1}^{n}\left(1+\frac{a_{k}b_{k}}{a_{0}b_{0}}\right)\label{eq:product def scalar n-dim}
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| \end{equation}
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| and the j$^{\text{th}}$ director component is
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| \begin{equation}
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| g_{j}=a_{0}b_{0}\prod_{k\neq j}^{n}\left(1+\frac{a_{k}b_{k}}{a_{0}b_{0}}\right)\left(\frac{a_{j}}{a_{0}}+\frac{b_{j}}{b_{0}}\right).\label{eq:product def director j n-dim}
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| \end{equation}
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| }The identity between the norm of a product and the product of the
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| norms described in \eqref{Conjugate-and-norm.} extended to $n$ dimensions
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| gives rise to Lagrange's and higher order identities\cite{Fernandezguasti12}.
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|
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| Complex numbers are to Euclidean geometry as hyperbolic numbers are
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| to Minkowski space-time special relativity \cite{catoni2008}. However,
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| just as it occurs in Euclidean space, a \{differentiable\} system
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| of hypercomplex numbers extended beyond the $1+1$ dimensional framework
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| has been elusive. It is thus natural to extend the 1+2 dimensional
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| hypercomplex system presented here to a deformed Lorentz - Minkowski
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| time-space.
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|
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| \bibliographystyle{unsrt}
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| \bibliography{/home/mfg/acad/ext/ref/mfg-arti,/home/mfg/acad/ext/ref/opt,/home/mfg/acad/ext/ref/phys,/home/mfg/acad/ext/ref/libros-doc}
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|
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| \end{document}
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|
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| --[[Usuario:Mfgwiki|mfg-wiki]] 22:16 11 ene 2010 (UTC) | | --[[Usuario:Mfgwiki|mfg-wiki]] 22:16 11 ene 2010 (UTC) |
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| [[Categoría:matematicas]] | | [[Categoría:matematicas]] |
