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
$ \ge2((\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_2|^2)
$ \because2次元の場合の式
$ =2|\bm e_0|^2|\bm e_1|^2|\bm e_2|^2((\hat{\bm e}_0\cdot\hat{\bm e}_1)(\hat{\bm e}_1\cdot\hat{\bm e}_2)(\hat{\bm e}_2\cdot\hat{\bm e}_0)-1)
$ =
負になる可能性もありそうだ