672 MOORE. 



is left invariant it is evident that r-* must vanish. That is 



Oimiki — Chmihi + cum^Jcz — azmoki = 



which requires that all the as vanish. However if we require that 

 r' = \r then we have r-$ = Xr which leads to the solutions 



a{hi^ih), h{h ± /A-4). 



Using these values for r we have 



r/ = (A-i + ihT)-'i! = hi + ih + imi{iki + iki) dt 



or -7^ = iniiih + iko). 



at 



Ti = (A'l — /A-o)-"^ = A-i + iki + i??ii(A-i — zX-2) (Zf 

 = — im\{k\ — iko). 



'dr2 



dt 

 Likewise for the other two we have 



—^ = iniiikz + iki), 

 dt 



— ^ = — imiikz — iki). 

 di 



Thus the points on ki + lAv and ^-3 + iki progress by the factors imi 

 and ivi2 while the point on ki — ik^ and kz — iki progress by the factors 

 — im\ and — iiiii. 

 If mi = 77?2 it is easily seen that any vector of the pencils 



(A-i + /A-o) + xa-3 + /A-4), (A-i - /A-o) + X(A-3 - /A'4) 



is left invariant. These vectors lie in a plane in which every vector 

 is a minimal vector. That is they lie in a plane which contains a 

 generator of the imaginary sphere at infinity. Any plane cutting 

 these two planes in a 1 -vector will be left invariant since it will contain 

 two invariant vectors. If jhi = —mo, the invariant pencils are 



(fci + ih) + X(A-3 - ^A-4), (A-i - ih^) + m(A-3 + iki). 



Each of these pencils lies in a plane cutting the imaginary sphere at 

 infinity in a generator and any plane cutting each of these planes in a 

 vector will be left invariant. We will however discuss these planes 

 from a different point of view. 



