Wedge積を帰納的に定義したい
$ {\bm v}_0\wedge{\bm v}_1\wedge\cdots\wedge{\bm v}_{n-1}=\sum_{i:\N\to\N}\epsilon_{{i_0}{i_1}\cdots{i_{n-1}}}{\bm v}_{i_0}{\bm v}_{i_1}\cdots{\bm v}_{i_{n-1}}
まずn=2は$ \bm{u}\wedge\bm{v}:=\bm{u}\bm{v}-\bm{v}\bm{u}でいい
n=3はどうなる？
$ \bm{u}_0\wedge(\bm{u}_1\wedge\bm{u}_2)=\bm{u}_0(\bm{u}_1\wedge\bm{u}_2)-(\bm{u}_1\wedge\bm{u}_2)\bm{u}_0
$ =\bm{u}_0\bm{u}_1\bm{u}_2-\bm{u}_0\bm{u}_2\bm{u}_1-\bm{u}_1\bm{u}_2\bm{u}_0+\bm{u}_2\bm{u}_1\bm{u}_0
$ \bm{u}_1\wedge(\bm{u}_0\wedge\bm{u}_2)=\bm{u}_1(\bm{u}_0\wedge\bm{u}_2)-(\bm{u}_0\wedge\bm{u}_2)\bm{u}_1
$ =\bm{u}_1\bm{u}_0\bm{u}_2-\bm{u}_1\bm{u}_2\bm{u}_0-\bm{u}_0\bm{u}_2\bm{u}_1+\bm{u}_2\bm{u}_0\bm{u}_1
そうはならなそうだ
成り立たないな
$ \underset{p}\bm{u}\wedge\underset{q}\bm{v}=(-1)^{pq}\underset{q}\bm{v}\wedge\underset{p}\bm{u}なのを思い出した