■基本単体の二面角(その418)
【8】E6格子の場合
P0(0,0,0,0,0,0,0)
P1(1,0,0,0,0,0,0)
P2(1,1/√3,0,0,0,0)
P3(1,1/√3,1/√3,0,0,0)
P4(1,1/√3,1/√3,1,0,0)
P5(1,1/√3,0,0,1/√3,0)
P6(1,1/√3,0,0,1/√3,1)
超平面をax+by+cz+dw+ev+fu=gとする.
[1]P1P2P3P4P5P6を通る超平面:x=1
[2]P0P2P3P4P5P6を通る超平面
g=0
a+b/√3=0,a=1,b=−√3
1−1+c/√3=0,c=0
1−1+c/√3+d=0,d=0
1−1+e/√3=0,e=0
1−1+e/√3+f=0,f=0
[3]P0P1P3P4P5P6を通る超平面
g=0,a=0
b/√3+c/√3=0,b=1,c=−1
b/√3+c/√3+d=0,d=0
b/√3+e/√3=0,e=−1
b/√3+e/√3+f=0,f=0
[4]P0P1P2P4P5P6を通る超平面
g=0,a=0,b=0
c/√3+d=0,c=√3,d=−1
e/√3=0,e=0
e/√3+f=0,f=0
[5]P0P1P2P3P5P6を通る超平面:w=0
[6]P0P1P2P3P4P6を通る超平面
g=0,a=0,b=0,c=0,d=0
e/√3+f=0,e=√3,f=−1
[7]P0P1P2P3P4P5を通る超平面:u=0
===================================
a=(1,0,0,0,0,0)
b=(1,−√3,0,0,0,0)
c=(0,1,−1,0,−1,0)
d=(0,0,√3,−1,0,0)
e=(0,0,0,1,0,0)
f=(0,0,0,0,√3,−1)
g=(0,0,0,0,0,1)
を正規化すると
a=(1,0,0,0,0,0)
b=(1/2,−√3/2,0,0,0,0)
c=(0,1/√3,−1/√3,0,−1/√3,0)
d=(0,0,√3/2,−1/2,0,0)
e=(0,0,0,1,0,0)
f=(0,0,0,0,0,√3/2,−1/2)
g=(0,0,0,0,0,1)
a・b=1/2
b・c=−1/2
c・d=−1/2
c・d=−1/2
d・e=−1/2
d・f=0
f・g=−1/2
===================================