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转到7个正射字。。。
John已经给出了典型正态((1,0,0,0,1,0,0))并列出了 14个顶点的点积为1。 我们可以把1放在任何坐标系中,然后取反,得到总共16个面。
对于形状的法线 $$\left(\pm\frac{1}{2},\pm\frac{1}},\fm\frac}{2{,\pm\frac{1},\ pm\frac}{2neneneep,0,0,0\right)$$ 坐标的排列有(2^4\cdot\binom{8}{4}=1120)的可能性。 例如, $$\left(\frac{1}{2},\frac}{1}},\frac{1}[2],\frac{1}[2},0,0,0\0\right)$$ 点积为1,具有以下14个顶点:
$$\开始{数组}{l} \左(\frac{1}{2},\frac}1}{2},\ frac{1}{2,\frac{1}{2},-\ frac}1}{2{,-\, -\压裂{1}{2},-\压裂{1{2}\右)\\ \左( \压裂{1}{2},压裂{1{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左( \压裂{1}{2},压裂{1{2}, -\压裂{1}{2},\压裂{1{2}\右)\\ \左( \压裂{1}{2},压裂{1{2}, \压裂{1}{2},-\压裂{1{2}\右)\\ \左( \frac{1}{2},\ frac{1}{2},\ frac{1}{2},\ frac{1}{2},\ frac{1}{2}, -\压裂{1}{2},\压裂{1{2}\右)\\ \左( \压裂{1}{2},压裂{1{2}, \压裂{1}{2},-\压裂{1{2}\右)\\ \左( \压裂{1}{2},压裂{1{2},压裂{1}{2}, -\压裂{1}{2},-\压裂{1{2}\右)\\ \左( \压裂{1}{2},压裂{1{2},压裂{1}{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左( 1,1,0,0,0,0,0 \右)\\ \左( 1、0、1、0,0、0、0,0\右)\\ \左( 1,0,0,1,0,0,0,0,0\右)\\ \左( 0,1,1,0,0,0,0,0\右)\\ \左( 0,1,0,1,0,0,0,0,0\右)\\ \左( 0,0,1,1,0,0,0 \右)\\ \结束{数组}$$
对于形状的法线 $$\left(\pm\frac{3}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4}\right)$$ 带有奇数个减号 坐标的排列有(8\cdot\left(\binom{8}{1}+\binom}{8}}{3}+\biom{8{5}+\Biom{8neneneep{7}\right)=1024)种可能性。 例如, $$\左(-\frac{3}{4},\frac}1}{4{,\frac{1}{4],\frac{1}}{4neneneep,\frac{1}{4},\frac:1}{4},\ frac{1{4}.,\frac/1}{3},\frac{1}4})$$ 点积为1,具有以下14个顶点:
$$\开始{array}{l} \左(-\frac{1}{2},-\frac{1}}{2{,\frac}1}{2],\frac{1{2}.,\frac, \压裂{1}{2},\压裂{1{2}\右)\\ \左( -\压裂{1}{2},压裂{1{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左( -\frac{1}{2},\ frac{1}{2},\ frac{1}{2},-\frac{1}{2},\ frac{1}{2},\ frac{1}{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左( -\压裂{1}{2},压裂{1{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左( -\压裂{1}{2},压裂{1{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左( -\压裂{1}{2},压裂{1{2},压裂{1}{2}, -\压裂{1}{2},\压裂{1{2}\右)\\ \左( -\压裂{1}{2},压裂{1{2},压裂{1}{2}, \压裂{1}{2},-\压裂{1{2}\右)\\ \左( -1,1,0,0,0,0,0,0\右)\\ \左( -1、0、1、0,0、0、0,0\右)\\ \左( -1,0,0,1,0,0,0,0,0\右)\\ \左( -1,0,0,0,1,0,0,0,0\右)\\ \左( -1,0,0,0,0,1,0,0\右)\\ \左( -1,0,0,0,0,0,1,0\右)\\ \左( -1,0,0,0,0,0,1\右)\\ \结束{数组}$$
这总共是16+1120+1024=2160个面。
我想你完全可以从对称性的角度来论证,这些由14个顶点组成的集合形成了7个正射曲面,因为约翰的例子很清楚 是 一个7-骨科。
从法线((\pm 2,\pm 1,\pm1,\pg 1,\fm 1,0,0,0)及其坐标的排列来看,有7个单形的(2^5\cdot 8\cdot\binom{7}{3}=8960)面。
从法线\(\ pm 1,\ pm 1,\ pm 1,\ pm 1,\ pm 1,\ pm 1,\ pm 1,\ pm 1)\) 奇数个负号 有7个简单的面(\binom{8}{1}+\binom}8}{3}+\biom{8{5}+\二进制{8}}{7}=128)。
从法线\((\pm\frac{3}{2},\pm\frac{3}}{2{,\pm\frac}3}{2],\pm\frac[1}{2neneneep,\pm\frac{1}{2neneneei,\pm\ frac{1}}{2},\ pm\frac{1}[2],\pm 奇数个负号 以及它们的坐标排列,有7个简单的面(\binom{8}{3}\cdot\left(\binom{8}}{1}+\binom}8}{3}+\biom{8{5}+\二进制{8}{7}\right)=7168)。
从法线\((\pm\frac{5}{2},\pm\frac{1}{2{,\pm\frac{1}},\ pm\frac{1}{2},\fm\frac}{1}[2},\tm\frac{1}{2neneneep,\pm\frac{1}{2}.,\pm\ frac{1{2}.\pm\fras{1}[2]) 偶数个减号 以及它们的坐标排列,有7个单形的(8\cdot左(\binom{8}{0}+\binom}8}{2}+\biom{8{4}+\bilom{8neneneep{6}+\二进制{8}}{8}\right)=1024)面。
这些计数的总数为17280。 这不是一个非常严格的枚举,因为我只是通过从\((2,1,1,1,1,0,0,0)\)开始的反复尝试找到它,并寻找具有相同平方范数8的其他向量。
我想我应该为每种法向量的一个例子展示7单纯形的8个顶点。 我已经为\((2,1,1,1,1,0,0,0)\)做过了,下面是其他三种类型的示例。
在每种情况下,法向量与顶点的点积是3。
对于\((1,-1,1,1,1,1,1)\),7-单形的8个顶点为:
$$ \开始{数组}{l} \left(-\frac{1}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左(\frac{1}{2},-\frac}1}{2],-\frac{1}}{2{,\frac[1}{2neneneep,\frac{1}{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左(\frac{1}{2},-\frac}1}{2],\frac[1}{2{,-\ frac{1'{2},\frac{1}}{2neneneep, \压裂{1}{2},\压裂{1{2}\右)\\ \左(\frac{1}{2},-\frac}1}{2],\frac[1}{2{,\frac{1{2},-\frac}1{2{2, \压裂{1}{2},\压裂{1{2}\右)\\ \左(\frac{1}{2},-\frac}1}{2],\frac[1}{2{,\frac{1{2},\frac{1}},\frac{1}[2}, \压裂{1}{2},\压裂{1{2}\右)\\ \left(\frac{1}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}, -\压裂{1}{2},\压裂{1{2}\右)\\ \左(\frac{1}{2},-\frac}1}{2],\frac[1}{2{,\frac{1{2},\frac{1}},\frac{1}[2], \压裂{1}{2},-\压裂{1{2}\右)\\ \左(\frac{1}{2},\frac}1}{2],\frac{1{2},\frac{1}}{2{,\ frac{1}{2}, \压裂{1}{2},\压裂{1{2}\右) \结束{数组} $$
对于\(左(-\frac{3}{2},\frac}3}{2},\ frac{3}{2,\ frac{3}},\frac{1}{2{,\frac{1{2},\frac{1}}{2),7个单纯形的8个顶点是:
$$ \开始{数组}{l} \左(-\frac{1}{2},\frac}1}{2],\frac{1{2},-\frac{1}{2}.,\frac}1}}, \压裂{1}{2},\压裂{1{2}\右)\\ \左(-\frac{1}{2},\frac}1}{2],\frac{1{2},\frac{1}}{2{,-\frac}1}},\ frac{1}{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \左图(-\frac{1}{2},\frac}1}{2],\frac{1{2},\frac{1}}{2{,\frac{1}[2],-\frac(1}{2)}, \压裂{1}{2},\压裂{1{2}\右)\\ \left(-\frac{1}{2},\ frac{1}{2},\ frac{1}{2},\ frac{1}{2},\ frac{1}{2}, -\压裂{1}{2},\压裂{1{2}\右)\\ \左图(-\frac{1}{2},\frac}1}{2],\frac{1{2},\frac{1}}{2{,\frac{1},\ frac{1}{2}, \压裂{1}{2},-\压裂{1{2}\右)\\ \left(-1,1,0,0,0,0,0\右)\\ \left(-1,0,1,0,0,0,0,0\右)\\ \左(0,1,1,0,0,0,0,0\右) \结束{数组} $$
对于\(左(\frac{5}{2},\frac}1}{2{,\frac{1}{2},\frac{1}}{2{,\frac{1{2},\frac 1}{2],\fric{1}[2],\fracc{1neneneep{2}\右),7个单纯形的8个顶点是:
$$ \开始{数组}{l} \左(\frac{1}{2},\frac}1}{2],\frac{1{2},\frac{1}}{2{,\ frac{1}{2}, \压裂{1}{2},\压裂{1{2}\右)\\ \left(1,1,0,0,0,0,0\右)\\ \左(1,0,1,0,0,0,0,0\右)\\ \左(1,0,0,1,0,0,0,0\右)\\ \左(1,0,0,0,1,0,0,0\右)\\ \左(1,0,0,00,0,1,0\右)\\ \左(1,0,0,0,0,0,1,0\右)\\ \left(1,0,0,0,0,0,0\右)\\ \结束{数组} $$
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