Complement of φ. Then we have the following relations between the components of the two plane:—
φ_{y z}^* = φ_{x l}, φ_{z x}^* = φ_{y l}, φ_{x y}^* = φ_{z l} φ_{z l}^* = φ_{y x} . . .
The proof of these assertions is as follows. Let u^*, v^* be the four vectors defining φ^*. Then we have the following relations:—
u_{x}^* u_{x} + u_{y}^* u_{y} + u_{z}^* u_{z} + u_{l}^* u_{l} = 0
u_{x}^* v_{x} + u_{y}^* v_{y} + u_{z}^* v_{z} + u_{l}^* v_{l} = 0
v_{x}^* u_{x} + v_{y}^* u_{y} + v_{z}^* u_{z} + v_{l}^* u_{l} = 0
v_{x}^* v_{x} + v_{y}^* v_{y} + v_{z}^* v_{z} + v_{l}^* v_{l} = 0
If we multiply these equations by v_{l}, u_{l}, v_{s}, and subtract the second from the first, the fourth from the third we obtain
u_{x}^* φ_{x l} + u_{y}^* φ_{y l} + u_{z}^* φ_{z l} = 0
v_{x}^* φ_{z l} + v_{y}^* φ_{y l} + v_{z}^* φ_{z l} = 0
multiplying these equations by v_{x}^* . u_{x}^*, or by v_{y}^* . u_{y}^*. we obtain
φ_{x z}^* φ_{x l} + φ_{y z}^* φ_{y l} = 0 and φ_{x y}^* φ_{x l} + φ_{z x}^* φ_{z l} = 0
from which we have
φ_{y z}^* : φ_{x y}^* : φ_{z x}^* = φ_{x l} : φ_{z l} : φ_{y l}
In a corresponding way we have
φ_{y z} : φ_{x y} : φ_{z x} = φ_{x l}^* : φ_{z l}^* : φ_{y l}^*.
ie.
φ_{i k}^* = λφ(_{i k})
when the subscript (ik) denotes the component of φ in the plane contained by the lines other than (ik). Therefore the theorem is proved.
We have
(φ φ*) = φ_{y z} φ_{y z}^* + . . .
= 2 (φ_{y z} φ_{z l} + . . .)
= 0
