Whitney消去双点

hbghlyj

当 $m = 1$ 时，8 字形用其引入的双点创建了一个圆盘，推过的动作相当简单Whitney消去双点可以表示为：
$ {\begin{cases}\alpha :\mathbb {R} ^{1}\to \mathbb {R} ^{2}\\\alpha (t)=\left({\frac {1}{1+t^{2}}},\ t-{\frac {2t}{1+t^{2}}}\right)\end{cases}} $
注意，如果将 $\alpha$ 视为映射到 $ \mathbb {R} ^{3} $，如下所示：
$ \alpha (t)=\left({\frac {1}{1+t^{2}}},\ t-{\frac {2t}{1+t^{2}}},0\right) $
那么可以通过以下方式将双点消去：
$ \beta (t,a)=\left({\frac {1}{(1+t^{2})(1+a^{2})}},\ t-{\frac {2t}{(1+t^{2})(1+a^{2})}},\ {\frac {ta}{(1+t^{2})(1+a^{2})}}\right). $
注意 $\beta(t, 0) = \alpha(t)$，并且当 $a ≠ 0$ 时，作为 $t$ 的函数，$\beta(t, a)$ 是一个嵌入。

当 $m = 2$ 时，
$ \alpha _{2}(t_{1},t_{2})=\left(\beta (t_{1},t_{2}),\ t_{2}\right)=\left({\frac {1}{(1+t_{1}^{2})(1+t_{2}^{2})}},\ t_{1}-{\frac {2t_{1}}{(1+t_{1}^{2})(1+t_{2}^{2})}},\ {\frac {t_{1}t_{2}}{(1+t_{1}^{2})(1+t_{2}^{2})}},\ t_{2}\right). $
这一过程最终引出了以下定义：
$ \alpha _{m}(t_{1},t_{2},\cdots ,t_{m})=\left({\frac {1}{u}},t_{1}-{\frac {2t_{1}}{u}},{\frac {t_{1}t_{2}}{u}},t_{2},{\frac {t_{1}t_{3}}{u}},t_{3},\cdots ,{\frac {t_{1}t_{m}}{u}},t_{m}\right), $
其中
$ u=(1+t_{1}^{2})(1+t_{2}^{2})\cdots (1+t_{m}^{2}). $

hbghlyj

celebratio.org/Whitney_H/article/220/

On the proof of the Whitney trick (from Scorpan's book)