計量行列の行列式
性質
2次元の場合常に非負
$ \because\det[\bm I]^{\sf EE}=|\bm e_0|^2|\bm e_1|^2-(\bm e_0\cdot\bm e_1)^2=|\bm e_0|^2|\bm e_1|^2(1-(\cos\theta)^2)=|\bm e_0|^2|\bm e_1|^2(\sin\theta)^2\ge0
$ \bm e_0\cdot\bm e_1=|\bm e_0||\bm e_1|\cos\thetaとした
3次元の場合は？
$ \det[\bm I]^{\sf EE}=|\bm e_0|^2|\bm e_1|^2|\bm e_2|^2+(\bm e_1\cdot\bm e_0)(\bm e_2\cdot\bm e_1)(\bm e_0\cdot\bm e_2)+(\bm e_2\cdot\bm e_0)(\bm e_0\cdot\bm e_1)(\bm e_1\cdot\bm e_2)
$ -(\bm e_1\cdot\bm e_0)(\bm e_0\cdot\bm e_1)(\bm e_2\cdot\bm e_2)-(\bm e_0\cdot\bm e_0)(\bm e_2\cdot\bm e_1)(\bm e_1\cdot\bm e_2)-(\bm e_2\cdot\bm e_0)(\bm e_1\cdot\bm e_1)(\bm e_0\cdot\bm e_2)
$ =|\bm e_0|^2|\bm e_1|^2|\bm e_2|^2+2(\bm e_1\cdot\bm e_0)(\bm e_2\cdot\bm e_1)(\bm e_0\cdot\bm e_2)-(\bm e_0\cdot\bm e_1)^2|\bm e_2|^2-(\bm e_1\cdot\bm e_2)^2|\bm e_0|^2-(\bm e_2\cdot\bm e_0)^2|\bm e_1|^2
$ =-2|\bm e_0|^2|\bm e_1|^2|\bm e_2|^2+2(\bm e_1\cdot\bm e_0)(\bm e_2\cdot\bm e_1)(\bm e_0\cdot\bm e_2)+(|\bm e_0|^2|\bm e_1|^2-(\bm e_0\cdot\bm e_1)^2)|\bm e_2|^2+(|\bm e_1|^2|\bm e_2|^2-(\bm e_1\cdot\bm e_2)^2)|\bm e_0|^2+(|\bm e_2|^2|\bm e_0|^2-(\bm e_2\cdot\bm e_0)^2)|\bm e_1|^2
$ =|\bm e_0|^2|\bm e_1|^2|\bm e_2|^2(-2+2(\hat\bm e_1\cdot\hat\bm e_0)(\hat\bm e_2\cdot\hat\bm e_1)(\hat\bm e_0\cdot\hat\bm e_2)+(1-(\hat\bm e_0\cdot\hat\bm e_1)^2)+(1-(\hat\bm e_1\cdot\hat\bm e_2)^2)+(1-(\hat\bm e_2\cdot\hat\bm e_0)^2))
$ =|\bm e_0|^2|\bm e_1|^2|\bm e_2|^2(-2+2\cos\theta_{01}\cos\theta_{12}\cos\theta_{20}+(\sin\theta_{01})^2+(\sin\theta_{12})^2+(\sin\theta_{20})^2)
$ \cos\theta_{ij}:=\hat\bm e_i\cdot\hat\bm e_jとした
$ \theta_{01},\theta_{12},\theta_{20}には制約条件があるはず
例えば$ \theta_{01}=\theta_{12}=\frac12\piなら$ \theta_{20}は任意の角度をとれる
$ \theta_{01}=\theta_{12}=\frac13\piだと、$ 0<\theta_{20}<\frac23\piである
$ \theta_{01}=\theta_{12}=\frac23\piも$ 0<\theta_{20}<\frac23\piである
つまり、$ 0<\theta_{20}<\min\{\theta_{01}+\theta_{12},2\pi-\theta_{01}-\theta_{12}\}か
$ \theta_{01},\theta_{12},\theta_{20}が全て鈍角だと、$ -2+2\cos\theta_{01}\cos\theta_{12}\cos\theta_{20}<0になる
負になる可能性もありそうだ