x=(x1,x2,x3),y=(y1,y2,y3)∈R3 olmak uzere Lie parantezini soyle tanimlayalim
[x,y]=x×y=|ijkx1x2x3y1y2y3|=(x2y3−x3y2)i−(x1y3−x3y1)j+(x1y2−x2y1)k
1)[x,y]=x×y=|ijkx1x2x3y1y2y3|=(x2y3−x3y2)i−(x1y3−x3y1)j+(x1y2−x2y1)k=−(y2x3−y3x2)i+(y1x3−y3x1)j−(y1x2−y2x1)k=−[(y2x3−y3x2)i−(y1x3−y3x1)j+(y1x2−y2x1)k]=−[y,x]
⟹[x,y] antisimetriktir
2)[λx,y]=(λx)×y=λ(x×y)=λ[x,y][x+y,z]=(x+y)×z=x×z+y×z=[x,z]+[y,z][z,x+y]=z×(x+y)=z×x+z×y=[z,x+[z,y]
⟹[x,y] iki lineerdir.
3)[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=x×(y×z)+y×(z×x)+z×(x×y)?=0
Oncelikle x×(y×z)=(x⋅z)y−(x⋅y)z ve a⋅b=b⋅a dir.
[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=x×(y×z)+y×(z×x)+z×(x×y)=(x⋅z)y−(x⋅y)z+(y⋅x)z−(y⋅z)x+(z⋅y)x−(z⋅x)y=0
⟹[x,y] Jacobi özdeşliği saglanir.