The hyperbolic space is defined by the constraint, $$t^2-\vec{x}^2=1 \quad \textbf{(1)},$$ where we think of $t$ as being a time coordinate and $\vec{x}=(x,y,z)$ as a spatial coordinate. This space is homogeneous and isotropic in the spatial coordinates which is what you are trying to prove.

**Isotropy**

Intuitively isotropy means that from the origin space looks the same in every direction. Formally this means that the space is invariant under rotations in space, since rotations can map any direction onto any other direction. By definition a rotations is a linear transformation of the spatial coordinates which leaves $\vec{x}^2 = x^2+y^2+z^2$ invariant or unchanged. The following three matrices are rotations about the $x$,$y$, and $z$ axes respectively. Every rotation can be written as a product of these matrices.

$$ R_x(\theta) =
\left[ \begin{array}
\ 1 & 0 & 0 \\
0 & \cos(\theta) & -\sin(\theta) \\
0 &\sin(\theta) & \cos(\theta) \\
\end{array} \right] $$

$$ R_y(\theta) =
\left[ \begin{array}
\ \cos(\theta) & 0 & -\sin(\theta) \\
0 & 1 & 0 \\
\sin(\theta) & 0& \cos(\theta) \\
\end{array} \right] $$

$$ R_z(\theta) =
\left[ \begin{array}
\ \cos(\theta) & -\sin(\theta) & 0 \\
\sin(\theta) & \cos(\theta) & 0 \\
0 & 0 & 1
\end{array} \right] $$

Where it is understood that these matrices are acting on the vector $\vec{x}=\left[x,y,z\right]^T$. In order to extend the definition of these operators to act on the three dimensional subspace of our space of four vectors $x^\mu=[t,x,y,z]^T$ we use the augmented matrix below which leave the time component alone,

$$
\left[ \begin{array}
\ 1 & \vec{0}^T \\
\vec{0} & R(\theta) \\
\end{array}\right]
$$

At this point it is elementary to see that this transformation leaves the quadratic form $t^2-\vec{x}^2$ invariant since the transformation doesn't change the value of $t$ or $\vec{x}^2$.

**Homogeneity**

We now wish to establish that the hyperbolic space is homogeneous. Intuitively this means that space is the same at every location. If we can show that space is isotropic when viewed from any location we will have established homogeneity. Mathematically this means that we need to show that every spatial coordinate can be mapped to the origin. So we need to find a transformation which leaves $t^2-\vec{x}^2$ invariant and maps $\vec{x}\rightarrow \vec{0}$.

To start suppose we have $\vec{x} = \vec{x}_0=(x_0,y_0,z_0)$. We start by rotating the vector so that it is parallel to the $x$ axis. The following transformation will accomplish this,

$$\frac{1}{\sqrt{x_0^2+y_0^2+z_0^2}\sqrt{y_0^2+z_0^2}}
\left[\begin{array}
\ x_0 & \sqrt{y_0^2+z_0^2} & 0 \\
-\sqrt{y_0^2+z_0^2} & x_0 & 0 \\
0 & 0 & 1
\end{array}\right]
\left[\begin{array}
\ 1 & 0 & 0 \\
0 & y_0 & z_0 \\
0 & -z_0 & y_0
\end{array}\right]
\left[\begin{array}
\ x_0 \\
y_0 \\
z_0 \\
\end{array}\right]
=
\left[\begin{array}
\ \sqrt{x_0^2+y_0^2+z_0^2} \\
0 \\
0 \\
\end{array}\right]
$$

Now we can map the spatial coordinate to zero by boosting. This will have the consequence of changing the $t$ coordinate of the four vector. Since the $y$ and $z$ coordinates are both $0$ due to the rotation we will only explicitly write the transformation in the two dimensions with nonzero entries $t$ and $x$.

$$
\left[ \begin{array} \
\cosh(\psi) & \sinh(\psi) \\
\sin(\psi) & \cosh(\psi)
\end{array} \right]
\left[ \begin{array} \
t_0 \\
\vert \vec{x}_0 \vert
\end{array} \right]
=
\left[ \begin{array} \
\cosh(\psi)t_0+\sinh(\psi)\vert \vec{x}_0 \vert \\
\sinh(\psi) t_0 + \cosh(\psi)\vert \vec{x}_0 \vert
\end{array} \right]
$$

We see that we can make the last spatial coordinate zero if $$\sinh(\psi) = -\frac{\vert \vec{x}_0 \vert}{\sqrt{t_0^2 - \vec{x}_0^2 }} \qquad \cosh(\psi) = \frac{t_0}{\sqrt{t_0^2-\vec{x}_0^2 }},$$
which is certainly possible for real values of $\psi$ so long as $\vec{x}_0^2 < t_0^2$ which is guaranteed by the condition $t^2=1+x^2$.

So we have shown that every point in space can be mapped to the origin.

This post imported from StackExchange Physics at 2014-06-18 14:05 (UCT), posted by SE-user Spencer