Post by Admin on Jan 23, 2017 15:23:06 GMT
Gary
Here is my take on this suggested exercise:
The half-space model of three-dimensional hyperbolic space
Let $K$ be a hemisphere centred at $q$ and orthogonal to the complex plane. Using (48) on page 324 we know that in the half-space model, a hemisphere orthogonal to the complex plane is an h-plane, and this leads us to suspect that $\mathcal{I}_K$ is reflection in this h-plane.
In symbols, $\mathfrak{R}_K(T)=\mathcal{I}_K(T)$, where $T=(X,Y,Z)$ is a point in hyperbolic half-space.
Let's be clear what we mean by reflection here. Think of the h-line $P$ through $T$ which cuts $K$ perpendicularly, say at $m$. We now define $\mathfrak{R}_K(T)$ to be the point on $P$ that is the same h-distance from $m$ as $T$.
Now think of figure 23b on page 304 as representing the vertical plane of the h-line $P$ described in the previous paragraph, replacing $z$ by $T$ and $\widetilde{z}$ by $\widetilde{T}=\mathcal{I}_K(T)$.
We know that every circle through $T$ and $\widetilde{T}$ is automatically orthogonal to $K$, and in particular the unique h-line through $T$ and $\widetilde{T}$ must be orthogonal to $K$ and hence it is the desired "$P$" of the previous paragraph.
Finally recall that from (14) on page 135 and the fact that a circle can be thought of as the intersection of two spheres, $\mathcal{I}_K$ maps $P$ into itself, swapping the segments $Tm$ and $\widetilde{T}m$. Thus, since we know (page 325) that $\mathcal{I}_K$ is a motion of hyperbolic space, these two h-line segments have equal h-length. This confirms that $\mathcal{I}_K$ is reflection in the h-plane $K$ and so we can generalise (38) as follows:
Inversion in a hemisphere $K$ orthogonal to the horizon is reflection $\mathfrak{R}_K$ of hyperbolic space in the h-plane $K$.
Vasco
Here is my take on this suggested exercise:
The half-space model of three-dimensional hyperbolic space
Let $K$ be a hemisphere centred at $q$ and orthogonal to the complex plane. Using (48) on page 324 we know that in the half-space model, a hemisphere orthogonal to the complex plane is an h-plane, and this leads us to suspect that $\mathcal{I}_K$ is reflection in this h-plane.
In symbols, $\mathfrak{R}_K(T)=\mathcal{I}_K(T)$, where $T=(X,Y,Z)$ is a point in hyperbolic half-space.
Let's be clear what we mean by reflection here. Think of the h-line $P$ through $T$ which cuts $K$ perpendicularly, say at $m$. We now define $\mathfrak{R}_K(T)$ to be the point on $P$ that is the same h-distance from $m$ as $T$.
Now think of figure 23b on page 304 as representing the vertical plane of the h-line $P$ described in the previous paragraph, replacing $z$ by $T$ and $\widetilde{z}$ by $\widetilde{T}=\mathcal{I}_K(T)$.
We know that every circle through $T$ and $\widetilde{T}$ is automatically orthogonal to $K$, and in particular the unique h-line through $T$ and $\widetilde{T}$ must be orthogonal to $K$ and hence it is the desired "$P$" of the previous paragraph.
Finally recall that from (14) on page 135 and the fact that a circle can be thought of as the intersection of two spheres, $\mathcal{I}_K$ maps $P$ into itself, swapping the segments $Tm$ and $\widetilde{T}m$. Thus, since we know (page 325) that $\mathcal{I}_K$ is a motion of hyperbolic space, these two h-line segments have equal h-length. This confirms that $\mathcal{I}_K$ is reflection in the h-plane $K$ and so we can generalise (38) as follows:
Inversion in a hemisphere $K$ orthogonal to the horizon is reflection $\mathfrak{R}_K$ of hyperbolic space in the h-plane $K$.
Vasco
