170 PHILLIPS AND MOORE. 



in the characteristic equation of 4>. For the discussion of this case 

 we may assume that there are no angles qa = 0. For such an angle 

 corresponds to an absolutely fixed plane. If there are several such 

 angles, the corresponding planes determine an absolutely fixed space 

 of ((>. Neither Mi nor M2 will have any planes lying in or cutting 

 this space. The discussion is then identical with that of motions in 

 the lower space perpendicular to this fixed space. 

 If there is only one angle equal to tt, we have 



Mi= irai-{- g2a2+ qnOin 



where ai, ao, •an are a set of completely . perpendicular planes. 

 This can be written 



il/i= x(ai+ a-2-\- 0:3+ • • ■ + oin) + (92— 7r)Q:2+ .... + (?«— 7r)a„. 

 It is easily seen that 



e 



I • ir(ai + 02'... + ail) -- T 



since it represents a rotation of 180° in each of the planes ai, ao. . .a;„. 

 Hence 



g/-M _ ^/•[(g2 — 7r)aj+ . . . . + (g — ;r)an] 



Similarly, we can write 



• M2= 7r/3i+^2|82+ .... +g„i3„ 



and 

 If now 



e'-M'-= — e^-[(92-'r)ft+.... + (5ii-T)/3„i 



(,1-Mi^ ^l-Mi 



we can apply the previous argument and so find 



(^0— 7r)a2 + + iqn— Tr)an = (q-2— t^)^-2 + -\-{qn— 7r)/3„, 



whence 



if 1- 7r(ai+ «2+ . . . + ««) = 3/2- 7r(i3i+ |32+ . . . . + /3„). 



