Consider $SL_2(\mathbb{R})$ as the set of $2\times 2$ real matrices of determinant $1$. Also consider $SO(2,1;\mathbb{R})$ as the group of real $3\times 3$ matrices of determinant one preserving the quadratic form of signature $(2,1)$. The two groups are isogeneous and there is a non-trivial $2$-to-$1$ group homomorphism $\phi:SL_2(\mathbb{R})\rightarrow SO(2,1;\mathbb{R})$.
My question is, given a matrix
$\begin{bmatrix} a&b\\ c&d \end{bmatrix}\in SL_2(\mathbb{R})$ ,
what matrix does $\phi$ map it to in $SO(2,1;\mathbb{R})$?
Best Answer
There is a related construction, where the quadratic form on $\mathbb R^3$ is not $x^2 + y^2 - z^2,$ rather $y^2 - zx.$ Let its Hessian matrix be $H.$ Then $H$ is real symmetric, eigenvalues $1,2,-1.$ The image of your matrix is $$ P = \left( \begin{array}{ccc} a^2 & 2ab & b^2 \\ ac & ad +bc & bd \\ c^2 & 2cd & d^2 \\ \end{array} \right) $$ which gives $$ P^T HP = H $$
Oh, if you are not demanding that $ad-bc=1,$ then $\det P = (ad-bc)^3$ and $P^THP = (ad-bc)^2 H$