Let $H:=\{(x_0,x_1,x_2)\in\mathbb{R}^3\mid -x_0^2+x_1^2+x_2^2=-1\}$ be the hyperbolic space with metric $g_{hip}$ induced by the Lorenz inner product $g_{Lor}=-dx_0^2+dx_1^2+dx_2^2$. Find a bijection between $H$ and $X:=\{M\in\mathbb{R}^{2\times 2}\mid M^t=M,\,\det(M)=1\}$ and show that $\text{SL}(2,\mathbb{R})$ acts on $H$ by isometries.

I've found the following correspondence: $(x_0,x_1,x_2)\in H\mapsto\left(\begin{array}{ll}(x_0+x_1) & x_2\\x_2 & (x_0-x_1)\end{array}\right)\in X$ and conversely $\left(\begin{array}{ll}a & b\\b & c\end{array}\right)\in X\mapsto\left(\frac{a+c}{2},\frac{a-c}{2},b\right)\in H$, which are mutually inverse.

I've also defined an action on $X$ given by: \begin{align*} \rho:\text{SL}(2,\mathbb{R})&\to \text{Bij}(X)\\ M&\mapsto (A\mapsto M^tAM) \end{align*} I've checked that $\rho$ is indeed a group action, but I don't know how to prove that this will be translated into an isometry of $H$.

Any suggestions?

  • $\begingroup$ Have you tried something? $\endgroup$ – Seub Sep 21 '18 at 18:55
  • $\begingroup$ @Seub, I tried to write $\rho(M)$ in terms of coordinates $x_0,x_1,x_2$ and show that $\rho(M)^*(-dx_0^2+dx_1^2+dx_2^2)=-dx_0^2+dx_1^2+dx_2^2$, but for some reason it didn't work (after lots of calculations, the coefficient of $dx_0^2$ wasn't $-1$, so I gave up) $\endgroup$ – rmdmc89 Sep 21 '18 at 21:42
  • $\begingroup$ the calculations were very long, so I thought maybe there's a better way $\endgroup$ – rmdmc89 Sep 21 '18 at 21:43
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    $\begingroup$ The direct, brute-force approach gives me the desired result. Likely you’re making a simple algebraic error somewhere along the way. $\endgroup$ – amd Sep 25 '18 at 23:22
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    $\begingroup$ I’m not sure that it’ll be any less tedious a computation, but another approach is to find the appropriate map $\Lambda$ from $\text{SL}(2,\mathbb R)$ to transformations of $H$ And show that $\Lambda(M)^tg\Lambda(M)=g$. All of these maps between spaces are linear, so you should be able to examine only basis vectors of $H$ and $X$. Endowing $X$ with a suitable inner product might help. $\endgroup$ – amd Sep 25 '18 at 23:59

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