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!


1 Answer 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.

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.

  • $\begingroup$ is that should be $\alpha \delta + \beta \gamma$ in 2nd row, 2nd column? $\endgroup$
    – Kerr
    Aug 4, 2016 at 3:21
  • $\begingroup$ @Jane That should do it. The book has these extra square root terms, so I had to typeset it here. If you wanted to get the quadratic form to be diagonal $(1,1,-1),$ you should be able to just write $CAC,$ where $C$ is the symmetric permutation matrix with ones on the backwards diagonal and zeros elsewhere. Oh, and this gives a group isomomorphism. $\endgroup$
    – Will Jagy
    Aug 4, 2016 at 3:30
  • 1
    $\begingroup$ Thank you! your matrix works very well for $z^2 - x^2 -y^2 =-1$ :) $\endgroup$
    – Kerr
    Aug 4, 2016 at 3:31

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