We see easily that this coincides with the usual rule for the vector-product; e. g., for j = x.
(AB_x) = A_y B_{x y} - A_z B_{z x}.
Correspondingly let us define in the four-dimensional case the product (Pf) of any four-vector P and the six-vector f. The j-component (j = x, y, z, or l) is given by
(Pf_{j}) = P_{x}f_{j x} + P_{y}f_{j y} + P_{w}f_{j z} + P_{z}f_{j l}
Each one of these components is obtained as the scalar product of P, and the vector f_j which is perpendicular to j-axis, and is obtained from f by the rule f_j = [(f_{j x}, f_{j y}, f_{j z}, f_{j l}) f_{j j} = 0.]
* * * * *
We can also find out here the geometrical significance of vectors of the third type, when f = φ, i.e., f represents only one plane.
We replace φ by the parallelogram defined by the two four-vectors U, V, and let us pass over to the conjugate plane φ^*, which is formed by the perpendicular four-vectors U^*, V^*. The components of (Pφ) are then equal to the 4 three-rowed under-determinants D_x D_y D_z D_l of the matrix
| P_x P_y P_z P_l |
| |
| U_x^* U_y^* U_z^* U_l^* |
| |
| V_x^* V_y^* V_z^* V_l^* |
Leaving aside the first column we obtain
D_x = P_y(U_z^* V_l^* - U_l^* V_z^*) + P_z(U_l^* V_y^* - U_y^* V_l^*)
+ P_l(U_y^* V_z^* - U_z^* V_y^*)
= P_y φ_{z y}^* + P_z^* φ_{l y} + P_l φ^*_{y z}._{y z}^* there]
= P_y φ_{x y} + P_z φ_{x z} + P_l φ_{x l},
which coincides with (Pφ_x) according to our definition.
