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Hyperbolic stereographic projection
相关: stereographic projection
Stereographic projection $σ:\mathbf H^n_R →\mathbf R^n$ and its inverse $σ^{−1}:\mathbf R^n →\mathbf H^n_R$ are given by
\begin{align*}
\sigma(\tau, \mathbf x) = \mathbf u &= \frac{R\mathbf x}{R + \tau},\\
\sigma^{-1}(\mathbf u) = (\tau, \mathbf x) &= \left(R\frac{R^2 + |u|^2}{R^2 - |u|^2}, \frac{2R^2\mathbf u}{R^2 - |u|^2}\right),
\end{align*}
where, for simplicity, $τ ≡ct$. The $(τ, \mathbf x)$ are coordinates on $\mathbf M^{n+1}$ and the $\mathbf u$ are coordinates on $\mathbf R^n$.
| 右图是$n=2$情形, 双曲面投影到单位圆盘 |
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$M$中的球(双曲面)的方程为
$$\mathbf H_R^n = \left\{\left(\tau, x^1, \ldots, x^n\right) \in M: -\tau^2 + \left(x^1\right)^2 + \cdots + \left(x^n\right)^2 = -R^2, \tau > 0\right\}$$
投影中心为 $S = (-R, 0, \ldots, 0)$.
如果$P = \left(\tau, x^1, \ldots, x^n\right) \in \mathbf H_R^n$，那么几何上很明显，$PS$与超平面 $\left\{\left(\tau, x^1, \ldots, x^n\right) \in M: \tau = 0\right\}$ 相交于一点，记为 $U = \left(0, u^1(P), \ldots, u^n(P)\right) \equiv (0, \mathbf u)$。我们有
\begin{align*}
S + \overrightarrow{SU} &= U \Rightarrow \overrightarrow{SU} = U - S,\\
S + \overrightarrow{SP} &= P \Rightarrow \overrightarrow{SP} = P-S
\end{align*}
即
\begin{align*}
\overrightarrow{SU} &= (0, \mathbf u) - (-R,\mathbf 0) = (R, \mathbf u),\\
\overrightarrow{SP} &= (\tau, \mathbf x) - (-R,\mathbf 0) = (\tau + R, \mathbf x).
\end{align*}
根据投影的定义，我们有
$$\overrightarrow{SU} = \lambda(\tau)\overrightarrow{SP}.$$
导出方程组
\begin{align}
R &= \lambda(\tau + R),\label3\\
\mathbf u &= \lambda \mathbf x.\label4
\end{align}
由第一个式子解出 $\lambda$ 得到 stereographic projection 的式子
$$\sigma(\tau, \mathbf x) = \mathbf u = \frac{R\mathbf x}{R + \tau}.$$
现在我们计算逆映射 $\sigma^{-1}(u) = (\tau, \mathbf x)$. 根据\eqref{3}\eqref{4}和
\begin{align*}
U &= (0, \mathbf u)\\
P &= (\tau(\mathbf u), \mathbf x(\mathbf u)).
\end{align*}
得到
\begin{align}
\tau &= \frac{R(1 - \lambda)}{\lambda},\label1\\
\mathbf x &= \frac{\mathbf u}{\lambda},\label2
\end{align}
但还没有结束: $\lambda$ 依赖于 $\mathbf u$.
点 $P$ 位于双曲面上的条件是
$$-\tau^2 + |\mathbf x|^2 = -R^2,$$
即
$$-\frac{R^2(1 - \lambda)^2}{\lambda^2}+\frac{|\mathbf u|^2}{\lambda^2}=-R^2,$$
即
$$-R^2(1 - \lambda)^2+|\mathbf u|^2=-R^2\lambda^2,$$
即
$$|\mathbf u|^2=R^2(1-2\lambda),$$
导出
$$\lambda = \frac{R^2 - |u|^2}{2R^2}.$$
代入\eqref{1}与\eqref{2}得
$$\sigma^{-1}(\mathbf u) = (\tau, \mathbf x) = \left(R\frac{R^2 + |u|^2}{R^2 - |u|^2}, \frac{2R^2\mathbf u}{R^2 - |u|^2}\right).$$ |
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