8
$\begingroup$

Let $$ \mathcal R=\{x=(x_1,x_2,x_3)\in\mathbb Z^3:x_1^2+x_2^2-x_3^2=-1\}. $$ The group $\Gamma= M_3(\mathbb Z)\cap O(2,1)$ acts on $\mathcal R$ by left multiplication. It's known that there is only one $\Gamma$-orbits in $\mathcal R$, i.e. $\Gamma \cdot e_3=\mathcal R$ where $e_3=(0,0,1)$.

Could anybody give me a proof of this fact?

Thanks.-.

[Comments: (i) $O(2,1)$ is the subgroup in $GL_3(\mathbb R)$ which preserves the form $x_1^2+x_2^2-x_3^2$, that is $$ O(2,1)=\{g\in GL_3(\mathbb R): g^t I_{2,1} g=I_{2,1}\}\qquad\textrm{where}\quad I_{2,1}= \begin{pmatrix}1&&\\&1&\\&&-1\end{pmatrix}. $$

(ii) $g(\mathcal R)\subset \mathcal R$ for any $g\in \Gamma$ because $g$ has integer coefficients and we can write $$ x_1^2+x_2^2-x_3^2=x^tI_{2,1}x, $$ then $$ (gx)^t I_{2,1} (gx)= x^t (g^tI_{2,1}g)x=x^tI_{2,1}x=-1. $$]

$\endgroup$
2
$\begingroup$

Do you know about Frink's paper? http://www.maa.org/sites/default/files/Orrin_Frink01279.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.