实心圆环

hbghlyj

这里说，$S^3=\{(z,w)\in  \mathbb{C}: |z|^2 +|w|^2=2\}$的子集$V_1=\{(z,w)\in  S^3: |z|\geq |w|\}$是一个实心圆环。为什么？

hbghlyj

$V_1$的边界$=\{(z,w)\in  S^3: |z|=|w|\}$，显然是一个面积为$(\pi({\tfrac {1}{\sqrt {2}}})^2)^2$的圆环表面。
${\tfrac {1}{\sqrt {2}}}S^{1}\times {\tfrac {1}{\sqrt {2}}}S^{1}=\left\{\left.{\tfrac {1}{\sqrt {2}}}(\cos \theta ,\sin \theta ,\cos \varphi ,\sin \varphi )\,\right|\,0\leq \theta <2\pi ,0\leq \varphi <2\pi \right\}.$


$V_1$可以写成
$\left\{\left.{\tfrac {1}{\sqrt {2}}}(|z|\cos \theta ,|z|\sin \theta ,|w|\cos \varphi ,|w|\sin \varphi )\,\right|\,0\leq \theta <2\pi ,0\leq \varphi <2\pi,1\ge|z|\ge|w|\ge0,|z|^2+|w|^2=1 \right\}.$

那怎样说明$V_1$是实心圆环？

Czhang271828

把条件
$$
1\geq |z|\geq |w|\geq 0\quad \text{and}\quad |z|^2+|w|^2=1
$$
替换成
$$
|z|=1\quad\text{and} \quad 1\geq |w|\geq 0
$$
即可. 这一变换是同胚的, 因为该变换的实质是
$$
(\cos\alpha,\sin\alpha)\mapsto (1,\tan\alpha),\quad \theta\in [0,\pi/4].
$$

hbghlyj

Czhang271828 发表于 2024-5-5 03:08
因为该变换的实质是
$$
(\cos\alpha,\sin\alpha)\mapsto (1,\tan\alpha),\quad \theta\in [0,\pi/4].
$$

谢谢！明白了。那个$\theta$是不是应该是$\alpha$？

hbghlyj

en.wikipedia.org/wiki/Clifford_torus#Alternative_derivation_using_complex_numbers

In fact, the Clifford torus divides this 3-sphere into two congruent solid tori (see Heegaard splitting^[2]).