I have asked this question few days ago in MathStackExchange but I got only one response which gave a partial answer to my question, so I decided to ask it here.


I am reading Kulkarni's "Proper action and Pseudo-Riemannian space forms" article. 


His work describes the action on the general space $S^{p,q}$, but I became interested in particular case:


one sheeted hyperboloid $S^{1,1}$ in $\mathbb{R}^3$ with equation
$$x^2+y^2-z^2=1.$$
$O(2,1)$ is defined to be a group of $Q$-orthogonal transformations preserving $S^{1,1}$. 


$Q$-orthogonality means preserving of the bilinear form 
$$b[(x\_1,y\_1,z\_1), (x\_2,y\_2,z\_2)] = x\_1x\_2+y\_1y\_2-z\_1z\_2.$$


$\textbf{My question is}$: are there any references/articles which describe the general form of elements of $O(2,1)$, or its generators, or decribe the finitely generated subgroups of $O(2,1)$. 


Thank you in advance!



orthogonal transformations of one sheeted hyperboloid $S^{1,1}$
This is from Magnus, *Noneuclidean Tesselations and Their Groups,* pages 123-124. In turn, this part is quoting fairly directly from [Fricke and Klein (1897), the first volume on automorphic forms, the volume on group theory](https://books.google.com/books?id=H5kLAAAAYAAJ&printsec=frontcover&dq=Vorlesungen%20%C3%BCber%20die%20Theorie%20der%20automorphen%20%20Functionen&hl=en&sa=X&ved=0CCAQ6AEwAGoVChMIscb4rNvYxgIVSCiICh3qMwQX#v=onepage&q=muesli&f=false). 


This is for the diagonal matrix $B$ with diagonal entries $(1,-1,-1),$ which is not the order you specify. Given real numbers $\alpha \delta - \beta \gamma = 1,$ let 
$$
A =
\left(
\begin{array}{ccc}
 \frac{1}{2} \left( \alpha^2 + \beta^2 + \gamma^2 + \delta^2 \right) & \alpha \beta + \gamma \delta & \frac{1}{2} \left( \alpha^2 - \beta^2 + \gamma^2 - \delta^2 \right) \\
\alpha \gamma + \beta \delta & \alpha \delta + \beta \gamma & \alpha \gamma - \beta \delta \\
 \frac{1}{2} \left( \alpha^2 + \beta^2 - \gamma^2 - \delta^2 \right) & \alpha \beta - \gamma \delta & \frac{1}{2} \left( \alpha^2 - \beta^2 - \gamma^2 + \delta^2 \right) 
\end{array}
\right)
$$
Then $$ A^T B A = B, $$
and all possible such $A$ with $\det A = 1$ are of this form. To get determinant $-1,$ just negate all entries of $A.$ 


This is the sort of setup that is common in differential geometry. For number theory of quadratic forms, it is cleaner to deal with the automorphism group of the indefinite form $y^2 - zx.$ If it is not clear yet, the number of shhets is defined by the nonzero constant to which the quadratic forms is set; the automorphism group does not change. 


