Negative curved spacetime Given a metric of the form $$ ds^2=c^2dt^2-a^2\left[d\chi^2+\frac{\sinh^2(\sqrt{-k}\chi)}{-k}d\Omega^2\right] $$
Where $d\Omega=d\theta^2+\sin^2(\theta)d\phi^2$, $k<0$
 and $a=a(t)$.
I came across the following question:
Show that the spatial part of the metric tensor can be written as a 3 dimensional
hyperboloid embedded into a 3+1 dimensional Minkowski space.
Now i actually have no idea where to begin. I suspect I'd have to find a map from $(\chi,\theta,\phi) \mapsto (x_1,x_2,x_3,x_4)$ such that $x_1^2+x_2^2+x_3^2-x_4^2=-1$ but i wouldn't know what it is, or how to get the equation. A hint or the solution would be much appreciated. 
 A: Welcome to stackexchange. 
Let's suppose the ambient metric on the $(3+1)$-dimensional Minkowski space $\mathbb{M}$ is given by 
$$ g = -dy_4 ^2 + dr^2 + r^2 d \Omega ^2 .$$
This just means we picked spherical coordinates for the spatial part. We want to find a submanifold $\mathbb{H} \subset \mathbb{M}$ which is a hyperboloid, such that the induced metric $\iota^* g$ on $\mathbb{H}$ is equivalent to the one from your question. To this end, consider 
$$ \mathbb{H} = \{y \in \mathbb{M} \quad | \quad y_4 ^2 - r^2 = \frac{1}{k} \}$$
where $r^2 = y_1 ^2 + y_2 ^2 + y_3 ^2$. 
It is easy to check that this submanifold is parametrized by $y_4 = \frac{\cosh(\sqrt{-k}\chi)}{\sqrt{-k}}$; $r =\frac{\sinh(\sqrt{-k}\chi)}{\sqrt{-k}}  $. You can also check that in terms of the new coordinate $\chi$, it holds that $-dy_4 ^2 + dr^2 = d\chi^2$.  The induced metric on $\mathbb{H}$ in terms of the coordinates $\chi , \Omega$ is thus given by $$ d\chi^2 + r^2 d \Omega^2 = d\chi^2 + \frac{\sinh^2(\sqrt{-k}\chi)}{-k}d\Omega^2. $$
