“We can also form a vectorial combination of a four-vector and a six-vector, giving us a vector of the third type. If the six-vector be of a special type, _i.e._, a piece of plane, then this vector of the third type denotes the parallelopiped formed of this four-vector and the complement of this piece of plane. In the general case, the product will be the geometric sum of two parallelopipeds, but it can always be represented by a four-vector of the 1st type. For two pieces of 3-space volumes can always be added together by the vectorial addition of their components. So by the addition of two 3-space volumes, we do not obtain a vector of a more general type, but one which can always be represented by a four-vector (loc. cit. p. 759). The state of affairs here is the same as in the ordinary vector calculus, where by the vector-multiplication of a vector of the first, and a vector of the second type (_i.e._, a polar vector), we obtain a vector of the first type (axial vector). The formal scheme of this multiplication is taken from the three-dimensional case. Let A = (A_{_x_}, A_{_y_}, A_{_z_}) denote a vector of the first type, B = (B_{_y z_}, B_{_z x_}, B_{_x y_}) denote a vector of the second type. From this last, let us form three special vectors of the first kind, namely— B_{_x_} = (B_{_x x_}, B_{_x y_}, B_{_x z_}) } B_{_y_} = (B_{_y x_}, B_{_y y_}, B_{_y z_}) } (B_{_i k_} = - B_{_k i_}, B_{_i i_} = 0). B_{_z_} = (B_{_z x_}, B_{_z y_}, B_{_z z_}) } Since B_{_j j_} is zero, B_{_j_} is perpendicular to the _j_-axis. The _j_-component of the vector-product of A and B is equivalent to the scalar product of A and B_{_j_}, _i.e._, (A B_{_j_},) = A_{_x_} B_{_j x_} + A_{_y_} B_{_j y_} + A_{_z_} B_{_j z_}. 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 (P_f_) of any four-vector P and the six-vector _f_. The _j_-component (_j_ = _x_, _y_, _z_, or _l_) is given by (P_f__{_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