ROTATIONS IN HYPERSPACE. 655 



II. 



2. Complex 2-vectors in 4-space. Let ^-i, ^-2, ^-3, ^4 be four mutu- 

 ally perpendicular unit vectors. Then any 1-vector can be expressed 

 as a linear function of these four and any 2-vector (simple or complex) 

 can be represented as a linear function of the six coordinate planes 



kij = ki>^kj. Thus 



(6) M = a^kn + 013^:13 + rtu^'u + a^ka + a24/c24 + a34^'34. 



From this equation we have at once, A comp'ex 2-mctor can he resolved 

 into the sum of two simp'e phmes Mi and Mi, one passing through an 

 arbitrary 1 -vector and the other lying in a 3-space perpendicuJar to it. 

 For let 



Ml = ai2A-i2 + aisA'is + Uuku 



This is a simple plane since it is the sum of three simple 2-vectors 

 having the vector A'l in common. As we could choose for A;i a unit 

 vector in any direction, this plane can be made to pass tlirough an 

 arbitrary 1-vector. Then let 



il/o = 023A'23 "I" (l2ik'2i + «34A'34- 



This is a plane since it is the sum of three simple plane vectors which 

 lie in the 3-space determined by ky, ks, ki. This 3-space is evidently 

 perpendicular to Ai since A'l • (A'2xA-3xA'4) = 0. (If we did not confine 

 ourselves to a rectangular set of axes this 3-space would not necessarily 

 be perpendicular to A"i). We can then write 



M = 3/1 -}- 3/2. 



The condition that M be a simple plane vector is 



MxM = 

 For, 



MxM = (Ml + M.)x(Mi + 3/2) = 23/1x3/2. 



In 4-space the cross product of two 2-vector3 is a scalar and if this 

 product vanishes it signifies that 3/i and 3/2 lie in a 3-space and con- 

 sequently 3/1 -{■ 3/2 can be expressed as a simple plane vector. 



A complex 2-vector can always be resohed into the sum of two simple 



