# 问题及解答

## 球面上等距同构是一非平凡 Clifford 平移的等价条件.

Posted by haifeng on 2021-06-24 09:24:52 last update 2021-06-24 09:24:52 | Edit | Answers (1)

References:

[1] V. V. Prasolov, V. M. Tikhomirov, Geometry.   Translations of Mathematical Monographs, Volume 200.

1

Posted by haifeng on 2021-06-24 10:37:57

\begin{aligned} x'_i&=x_i\cos\alpha_i+y_i\sin\alpha_i,\\ y'_i&=-x_i\sin\alpha_i+y_i\cos\alpha_i,\\ z'_i&=\pm z_i. \end{aligned}

$2\Bigl(\sum_{i=1}^{k}(x_i^2+y_i^2)(1-\cos\alpha_i)+\sum_{j=1}^{s}\varepsilon_j z_j^2\Bigr),$

$\begin{split} d^2&=\sum_{i=1}^{k}\bigl[(x'_i-x_i)^2+(y'_i-y_i)^2\bigr]+\sum_{i=1}^{s}(z'_i-z_i)^2\\ &=\sum_{i=1}^{k}\Bigl[(x_i\cos\alpha_i+y_i\sin\alpha_i-x_i)^2+(-x_i\sin\alpha_i+y_i\cos\alpha_i-y_i)^2\Bigr]+\sum_{i=1}^{s}(\pm z_i-z_i)^2\\ &=\sum_{i=1}^{k}\Bigl[(\cos\alpha_i-1)^2 x_i^2+2(\cos\alpha_i-1)x_i y_i\sin\alpha_i+y_i^2\sin^2\alpha_i+x_i^2\sin^2\alpha_i-2x_i\sin\alpha_i(\cos\alpha_i-1)y_i+(\cos\alpha_i-1)^2 y_i^2\Bigr]+2\sum_{j=1}^{s}\varepsilon_j z_j^2\\ &=\sum_{i=1}^{k}\Bigl[x_i^2(\cos^2\alpha_i-2\cos\alpha_i+1+\sin^2\alpha_i)+y_i^2(\sin^2\alpha_i+\cos^2\alpha_i-2\cos\alpha_i+1)\Bigr]+2\sum_{j=1}^{s}\varepsilon_j z_j^2\\ &=\sum_{i=1}^{k}\Bigl[2(1-\cos\alpha_i)x_i^2+2(1-\cos\alpha_i)y_i^2\Bigr]+2\sum_{j=1}^{s}\varepsilon_j z_j^2\\ &=2\biggl[\sum_{i=1}^{k}\Bigl((x_i^2+y_i^2)(1-\cos\alpha_i)\Bigr)+\sum_{j=1}^{s}\varepsilon_j z_j^2\biggr] \end{split}$

$\sum_{i=1}^{k}(x_i^2+y_i^2)+\sum_{j=1}^{s}z_j^2=1$

• (i) $k=0$, 且 $\varepsilon_1=\cdots=\varepsilon_s=\pm 1$;
• (ii) $s=0$, 且 $\cos\alpha_1=\cdots=\cos\alpha_k$.