結局、シュレーフリ記号と２面角から3行目を決めて(2,0)=2とすればよい

ユークリッド空間の基本単体では

(sinα)^2(sinγ)^2-(cosβ)^2=0

sinαsinγ-cosβ=0

(tanα)^2(tanγ)^2-(cosβ)^2/(cosα)^2(cosγ)^2=0

　が導き出せればよい

4次元の場合はどれくらい複雑になるのだろうか？

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(0,2)(1,3)=(secβ)^2

(1,3)(2,4)=(secγ)^2

(1,3)=(secβ)(secγ)

(0,2)=(secβ)(cosγ)

(2,4)=(secγ)(cosβ)

(-1,1)(0,2)=(secα)^2

(2,4)(3,5)=(secδ)^2

(-1,1)=(secα)^2cosβsecγ

(3,5)=(secδ)^2secβcosγ

(-1,2)=(tanα)^2

(0,3)=(tanβ)^2

(1,4)=(tanγ)^2

(2,5)=(tanδ)^2

(tanα)^2(tanβ)^2-(secβ)(cosγ)(-1,3)=1

(-1,3)={(tanα)^2(tanβ)^2-1}/(secβ)(cosγ)

={(sinα)^2(sinβ)^2-(cosα)^2(cosβ)^2}/(cosα)^2cosβcosγ

(tanβ)^2(tanγ)^2-(secβ)(secγ)(0,4)=1

(0,4)={(tanβ)^2(tanγ)^2-1}/(secβ)(secγ)

={(sinβ)^2(sinγ)^2-(cosβ)^2(cosγ)^2}/cosβcosγ

(tanγ)^2(tanδ)^2-(secγ)(cosβ)(1,5)=1

(1,5)={(tanγ)^2(tanδ)^2-1}/(secγ)(cosβ)

={(sinγ)^2(sinδ)^2-(cosγ)^2(cosδ)^2}/(cosδ)^2cosβcosγ

(-1,3)(0,4)-(tanβ)^2(-1,4)=1

{(sinα)^2(sinβ)^2-(cosα)^2(cosβ)^2}{(sinβ)^2(sinγ)^2-(cosβ)^2(cosγ)^2}/(cosα)^2(cosβ)^2(cosγ)^2-(tanβ)^2(-1,4)=1

(0,4)(1,5)-(tanγ)^2(0,5)=1

{(sinγ)^2(sinδ)^2-(cosγ)^2(cosδ)^2}{(sinβ)^2(sinγ)^2-(cosβ)^2(cosγ)^2}/(cosδ)^2(cosβ)^2(cosγ)^2-(tanγ)^2(0,5)=1

(-1,4)(0,5)=1

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(-1,4)(0,5)-(0,4)(-1,5)=1

(-1,5)=[(-1,4)(0,5)-1}/(0,4)

(-1,4)={{(sinα)^2(sinβ)^2-(cosα)^2(cosβ)^2}{(sinβ)^2(sinγ)^2-(cosβ)^2(cosγ)^2}/(cosα)^2(cosβ)^2(cosγ)^2-1}/(tanβ)^2

={{(tanα)^2(tanβ)^2-1}{(tanβ)^2(tanγ)^2-1}(cosβ)^2-1}/(tanβ)^2

=(cosβ)^2{{(tanα)^2(tanβ)^2(tanγ)^2-(tanα)^2-(tanγ)^2 -1}

(0,5)={{(sinγ)^2(sinδ)^2-(cosγ)^2(cosδ)^2}{(sinβ)^2(sinγ)^2-(cosβ)^2(cosγ)^2}/(cosδ)^2(cosβ)^2(cosγ)^2-1}/(tanγ)^2

={{(tanβ)^2(tanγ)^2-1}{(tanγ)^2(tanδ)^2-1}(cosγ)^2-1}/(tanγ)^2

=(cosγ)^2{{(tanδ)^2(tanβ)^2(tanγ)^2-(tanδ)^2-(tanβ)^2 -1}

(-1,4)(0,5)=1

{{(tanα)^2(tanβ)^2(tanγ)^2-(tanα)^2-(tanγ)^2 -1}{{(tanδ)^2(tanβ)^2(tanγ)^2-(tanδ)^2-(tanβ)^2 -1}=(secβ)^2(secγ)^2

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