I am given coordinates on the Minkowski space:
$Z_{0}=\frac{t^2-r^2-l^2}{2t}$,
$Z_{i}=l\frac{x_{i}}{t}$ with $i=1,2,3$,
$Z_{4}=\frac{t^2-r^2-l^2}{2t}$
where we have $r^2=x_{1}^2+x_{2}^2+x_{3}^2$.
Now I am wondering how to verify that for the metric we have:
$ds^2=\frac{l^2}{t^2}(-dt^2+dx_{1}^2+dx_{2}^2+dx_{3}^2+dx_{4}^2)$.
All I have been able to calculate is that we have $-Z_{0}^2+\sum_{i=1}^{4}Z_{i}^2=\frac{l^2\cdot r^2}{t^2}$.
Furthermore I know that we are able to represent the metric by a matrix:
$g_{\mu \nu}=\frac{l^2}{t^2} diag(-1,1,1,1,1)$, but sadly I do not have an idea how to verify the form above...
I am thankful for every answer and comment you leave! Thank you in advance!
The Minkowski metric is given by $-dZ_0^2 + \sum_{i=1}^4 dZ_i^2$. We want to pull this back by the parametrization you've supplied. (I'm assuming that $l$ is constant.) Well, \begin{align*} dZ_0 &= \frac12(1-\frac{l^2}{t^2})dt - \frac1t(x_1\,dx_1+x_2\,dx_2+x_3\,dx_3) \\ dZ_i &= l\left(-\frac{x_i}{t^2}dt + \frac1t\,dx_i\right), \quad i=1,2,3 \\ dZ_4 &= dZ_0 \end{align*} So, of course, when we calculate the pullback, $-dZ_0^2 + dZ_4^2 = 0$, and we're left with $$\sum_{i=1}^3 dZ_i^2 = l^2\left(\frac{r^2}{t^4}dt^2 + \frac1{t^2}\sum_{i=1}^3 dx_i^2-\frac{x_i(dt\,dx_i + dx_i\,dt)}{t^3}\right).$$ I'm assuming there's an error in your question. In particular, how can $dx_4$ appear? Is it possible that $r^2 = x_1^2+\dots+x_4^2$? I also think that if you're talking about "half of Minkowski space," you certainly cannot have $Z_0=Z_4$. Indeed, what do you even mean by this phrase? Please double-check everything.