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In this paper, page $6$, the authors state the following:

The translations of the hyperbolic plane are defined as products of two central symmetries; the set of hyperbolic translations forms a sharply transitive set on the hyperbolic plane, the associated loop is the classical simple Bruck loop.

I would like to have a refence to study about the boldfaced sentences in the text above.

I would appreciate your help.

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These are deep waters.

Central symmetries are defined on the first page of T. Banakh, A. Dudko, D. Repovš

http://new.math.uiuc.edu/math402/public/models/htranslation.html

http://en.wikipedia.org/wiki/Square_root_of_a_matrix#Square_roots_of_positive_operators

http://en.wikipedia.org/wiki/Bol_loop

Using the Poincare disk model, a hyperbolic translation is the Möbius transformation given by matrix $$ A \; = \; \left( \begin{array}{rr} 1 & \alpha \\ \bar{\alpha} & 1 \end{array} \right) , $$ or $$ T_\alpha(z) = \frac{z + \alpha}{\bar{\alpha}z + 1 } \; \; , $$ where $\alpha$ is a complex number with $|\alpha| < 1.$ The transformation does take the unit circle to itself as along as $\alpha$ is not itself on the unit circle. However, we also need $T_\alpha(0)$ to be inside the disk, and $T_\alpha(0) = \alpha.$

The matrix for $T_\alpha \circ T_\beta$ has matrix $$ \left( \begin{array}{rr} 1 + \alpha \bar{\beta} & \alpha + \beta \\ \bar{\alpha} + \bar{\beta} & 1 + \bar{\alpha} \beta \end{array} \right) . $$ This is simply not Hermitian, and other things go wrong.

Given two positive Hermitian matrices,
$$ A \; = \; \left( \begin{array}{rr} 1 & \alpha \\ \bar{\alpha} & 1 \end{array} \right) , $$ and $$ B \; = \; \left( \begin{array}{rr} 1 & \beta \\ \bar{\beta} & 1 \end{array} \right) , $$ as you see above $AB$ or $BA$ are not Hermitian, so the classical Bruck loop operation is $$ \sqrt{B A^2 B},$$ where the square root is the (unique) positive Hermitian matrix $H$ such that $ H^2 = B A^2 B.$ So, the operation is not commutative or associative, but it satisfies a Bol identity and behaves well as far as inverses.

EDIT, Saturday, March 24, 12:23 Pacific time: got it, very hard. We get, with the above $A,B$ as written, $$ B A^2 B = \left( \begin{array}{rr} | 1 + \bar{\alpha} \beta |^2 + |\alpha + \beta|^2 & 2 (\alpha + \beta) (1 + \bar{\alpha} \beta) \\ 2 (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta}) & | 1 + \bar{\alpha} \beta |^2 + |\alpha + \beta|^2 \end{array} \right). $$ Now, as matrices, we get $$ \sqrt{B A^2 B} = \left( \begin{array}{cc} | 1 + \bar{\alpha} \beta | & \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ | 1 + \bar{\alpha} \beta |} \\ \frac{ (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{ | 1 + \bar{\alpha} \beta |} & | 1 + \bar{\alpha} \beta | \end{array} \right). $$ However, as a Möbius transformation, we may divide all four entries by the real number $ | 1 + \bar{\alpha} \beta |,$ and call that transformation $A * B.$ That is, $$ A * B = \left( \begin{array}{cc} 1 & \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ | 1 + \bar{\alpha} \beta |^2} \\ \frac{ (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{ | 1 + \bar{\alpha} \beta |^2} & 1 \end{array} \right) $$ or $$ A * B = \left( \begin{array}{cc} 1 & \frac{ (\alpha + \beta) (1 + \bar{\alpha} \beta)}{ ( 1 + \alpha \bar{\beta} ) ( 1 + \bar{\alpha} \beta ) } \\ \frac{ (\bar{\alpha} + \bar{\beta}) (1 + \alpha \bar{\beta})}{( 1 + \bar{\alpha} \beta ) ( 1 + \alpha \bar{\beta} ) } & 1 \end{array} \right) $$ or $$ A * B = \left( \begin{array}{cc} 1 & \frac{ \alpha + \beta }{ 1 + \alpha \bar{\beta} } \\ \frac{ \bar{\alpha} + \bar{\beta}}{ 1 + \bar{\alpha} \beta } & 1 \end{array} \right). $$ We finally have what Grishkov and Nagy write, using Möbius transformations with both diagonal elements equal to $1$ and the matrix Hermitian in any case, everything is determined by a complex number of modulus less than $1$ in the upper right corner. Put together, we have $$ A \; = \; \left( \begin{array}{rr} 1 & \alpha \\ \bar{\alpha} & 1 \end{array} \right) , $$ and $$ B \; = \; \left( \begin{array}{rr} 1 & \beta \\ \bar{\beta} & 1 \end{array} \right) , $$ then $$ A * B = \left( \begin{array}{cc} 1 & \frac{ \alpha + \beta }{ 1 + \alpha \bar{\beta} } \\ \frac{ \bar{\alpha} + \bar{\beta}}{ 1 + \bar{\alpha} \beta } & 1 \end{array} \right). $$

With $x= \alpha, y = \beta,$ we see the Grishkov and Nagy formula at the bottom of page 6, $$ x \cdot y = \frac{x + y}{1 + x \bar{y}} = \frac{ \alpha + \beta }{ 1 + \alpha \bar{\beta} }.$$

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