kaleidoscopes, p295

の大域・局所問題を計算する．

{3,3,4}(1,0,0,0)・・・(8,24,32,16),(1,6,12,8)

{3,3,4}(0,1,0,0)・・・(24,96,96,24),(1,8,12,6)

{3,3,4}(0,0,1,0)・・・(32,96,88,24),(1,6,9,5)

{3,3,4}(0.0.0,1)・・・(16,32,24,8),(1,4,6,4)

{3,3,4}(1,1,0,0)・・・(48,120,96,24),(1,5,8,5)

{3,3,4}(1,0,1,0)・・・(96,288,240,48),(1,6,9,5)

{3,3,4}(1,0,0,1)・・・(64,192,208,80),(1,6,12,8)

{3,3,4}(0,1,1,0)・・・(96,192,120,24),(1,4,6,4)

{3,3,4}(0,1,0,1)・・・(96,288,248,56),(1,6,9,5)

{3,3,4}(0,0,1,1)・・・(64,128,88,24),(1,4,6,4)

{3,3,4}(1,1,1,0)・・・(192,384,240,48),(1,4,6,4)

{3,3,4}(1,1,0,1)・・・(192,480,368,80),(1,5,8,5)

{3,3,4}(1,0,1,1)・・・(192,480,368,80),(1,5,8,5)

{3,3,4}(0,1,1,1)・・・(192,384,248,56),(1,4,6,4)

{3,3,4}(1,1,1,1)・・・(384,768,464,80),(1,4,6,4)

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{3,3,4}(0,1,0,0)・・・(24,96,96,24),(1,8,12,6)

6 1　　

12 0　　

8 0 1

1 0 0 1

0 0 0 0 1

これにD4(8,-24,32,16,1)をかければOK

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1 1

4 0　　

4 0 1

1 0 0 1

0 0 0 0 1

これに(2,-1,3,3,1)をかけてもNG

これに(2,-1,4,4,1)をかければOK

最後の二重節点の位置は｛３，３｝（０１０）である．

｛３，３｝（０１０）の局所幾何（１，４，４，１）より

ｍ＝（２，-１，４，４，１）としてみる．

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