O(1,1)

Post

O(1,1)

In Math on June 29, 2010 by hoxide

This is too simple to write in a notes, but it is still interesting.
O(p,q) preserve the quadratic form with signature (p,q).

Then what is O(1,1). The conclusion is that $O(1,1)\cong \mathbb{R}^+\rtimes ( \mathbb{Z}_2\times \mathbb{Z}_2)$

To see this, it is clear that
$\begin{pmatrix}\pm 1 & 0 \\ 0& \pm 1\end{pmatrix}$
form a subgroup $K$ of O(1,1) isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$

$\lambda \in\mathbb{R}^+$ correspond to matrix
$\frac{1}{2} \begin{pmatrix}\lambda+\lambda^{-1} &\lambda-\lambda^{-1} \\ \lambda-\lambda^{-1}& \lambda+\lambda^{-1}\end{pmatrix}$
which form subgroup $N$ in O(1,1).
$K$ act on $N$ by $\lambda \mapsto \lambda^{\det(g)}$ .

The way to see this is by the consideration of nill-cone In fact, O(p,q) preserve the nill-cone. For O(1,1), the nill-cone is $x^2-y^2=0$ , just two lines. Now let $\lambda$ act by scaler on the line $x=y$ , then the action on the other line has no choice to be scaler by $\lambda^{-1}$ . $K$ just permute the lines or inverse the direction of line. Or we can say $O(1,1) \cong \mathbb{R}^\times \rtimes \mathbb{Z}_2$ by choice of the permutation between lines.