Is $PGL(2,\mathbb R)$ isomorphic to $SO(1,2)$

finite-groupsgroup-isomorphismgroup-theorylie-groupsmatrices

Consider the following representation $\rho:GL(2,\mathbb R)\to GL(3,\mathbb R):G\mapsto \hat{G}$ where

$G=\begin{pmatrix}
\alpha&\beta\\
\gamma&\delta
\end{pmatrix}\in GL(2,\mathbb R)$

and

$\hat{G}:=
\frac{1}{\Delta}
\begin{pmatrix}
\frac{1}{2}\big(\alpha^2+\beta^2+\gamma^2+\delta^2\big)&\frac{\sqrt{2}}{2}(\alpha\gamma+\beta\delta)&-\frac{1}{2}\big(\alpha^2+\beta^2-\gamma^2-\delta^2\big)\\
\sqrt{2}\, (\alpha\beta+\gamma\delta)&\alpha\delta+\beta\gamma&-\sqrt{2}\, (\alpha\beta-\gamma\delta)\\
-\frac{1}{2}\big(\alpha^2-\beta^2+\gamma^2-\delta^2\big)&-\frac{\sqrt{2}}{2}(\alpha\gamma-\beta\delta)&\frac{1}{2}\big(\alpha^2-\beta^2-\gamma^2+\delta^2\big)
\end{pmatrix}\in GL(3,\mathbb R)$
with $\Delta=\det G=\alpha\delta-\beta\gamma$.

The kernel of $\rho$ is the normal subgroup of nonzero scalar transformations of $\mathbb R^2$, namely $Z=\{\lambda I_2|\lambda\in\mathbb R^\times\}$.

Hence, $\rho$ induces a faithful representation $PGL(2,\mathbb R)\to GL(3,\mathbb R)$ where $PGL(2,\mathbb R)$ is the projective general linear group $PGL(2,\mathbb R)=GL(2,\mathbb R)/Z$.

Note furthermore that $\det \hat{G}=1$ and $\hat{G}^T\eta\hat{G}=\eta$ where $\eta=\begin{pmatrix}
-2&0&0\\
0&1&0\\
0&0&2
\end{pmatrix}$
, hence $\hat{G}\in SO(1,2)$ [since $\eta$ has signature $(1,2)$] where $SO(1,2)$ stands for the proper Lorentz group in $2+1$ dimensions.

Whenever $G\in SL(2,\mathbb R)$, $\hat{G}{}^0{}_{0}=\frac{1}{2}\big(\alpha^2+\beta^2+\gamma^2+\delta^2\big)>0$, so that $\hat{G}\in SO^+(1,2)$ with $SO^+(1,2)$ the proper orthochronous Lorentz group in $2+1$ dimensions, so that $\rho$ reduces to the well-known isomorphism $PSL(2,\mathbb R)\to SO^+(1,2)$, where $PSL(2,\mathbb R)=SL(2,\mathbb R)/\{-1,1\}$ is the projective special linear group in 2 dimensions.

Question:

Does $\rho$ induce an isomorphism $PGL(2,\mathbb R)\to SO(1,2)$?

If yes, are there some references discussing this isomorphism?

Best Answer

Yes, the $\rho$ you constructed is an isomorphism and it is also true when $\Bbb R$ is replaced with any field of characteristic different from $2$.

This is in fact a well known result which can be proved using quaternion algebras. A reference is Arithmetique des algebres de quaternions by Marie-France Vigneras, Chapitre I, theoreme 3.4, 1). The book is in French though.

In short, for $K$ a field of characteristic different from $2$, we consider the quaternion algebra $Q = \operatorname M_2(K)$ and its trace zero space $Q^0$.

The reduced norm map $n\colon Q^0 \rightarrow K$ is a quadratic form on $Q^0$. Under the basis $$i = \begin{pmatrix} & 1\\ 1 & \end{pmatrix}, j = \begin{pmatrix}-1 & \\ & 1\end{pmatrix}, k = \begin{pmatrix}& 1\\-1&\end{pmatrix}$$ of $Q^0$, the quadratic form $n$ has matrix $\begin{pmatrix}-1 & &\\& -1 &\\& & 1\end{pmatrix}$.

Therefore $\operatorname{SO}_{1, 2}(K)$ can be identified with the group of isometries of the quadratic space $(Q^0, n)$ with determinant $1$, which is again isomorphic to $Q^\times / K^\times$, by a general result on quaternion algebras.

This last group is nothing but $\operatorname{GL}_2(K)/K^\times$, which is $\operatorname{PGL}_2(K)$.