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[几何] 保持连通的情况下最多可以移除的 $k$ 维曲线数

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hbghlyj posted 2024-8-28 23:04 |Read mode
$b_k$ 是对象保持连通的情况下最多可以移除的 $k$ 维曲线数。
Torus_cycles[1].png
例如,在移除两个一维曲线(赤道和子午线)后,圆环仍然保持连通,因此 $b_1 = 2$.

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original poster hbghlyj posted 2024-8-28 23:12
Last edited by hbghlyj 2024-10-5 13:33Mathematica
$$b_0=1,b_1=0,b_2=0$$
$$b_0=1,b_1=1$$
為什麼呢

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original poster hbghlyj posted 2024-8-29 09:07
hbghlyj 发表于 2024-8-28 15:12
ResourceFunction["BettiNumbers"][Simplex[{1, 2, 3}]]$=\{1,0,0\}$
為什麼呢
按照mit18_900s23_lec29.pdf的公式:$\begin{aligned} & b_0=n_0-\operatorname{rank}\left(D_1\right) \\ & b_1=n_1-\operatorname{rank}\left(D_1\right)-\operatorname{rank}\left(D_2\right) \\ & b_2=n_2-\operatorname{rank}\left(D_2\right)\end{aligned}$
对于Simplex[{1, 2, 3}]有
$D_1=\pmatrix{1&1\\-1&&1\\&-1&-1}$
$D_2=\pmatrix{1\\1\\-1}$
$n_0=3$
$n_1=3$
$\operatorname{rank}D_1=2$
$n_2=1$
$\operatorname{rank}D_2=1$
$b_0=n_0-\operatorname{rank}D_1=1$
$b_1=n_1-\operatorname{rank}D_1-\operatorname{rank}D_2=0$
$b_2=n_2-\operatorname{rank}D_2=0$
即$\{1,0,0\}$

那这个公式又是為什麼呢

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