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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 $$ (gx)^t I_{2,1} (gx)= x^t (g^tI_{2,1}g)x=x^tI_{2,1}x=-1. $$]

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1 Answer 1

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

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