188 PHILLIPS AND MOORE. 



are commutative. Hence 



which is the resolution required. Conversely, this can be done, 

 except for sign, in only one way. For suppose 



4>\ and 03 being equiangular rotations of one type, 02 and 4>i of the 

 opposite type. Then 



03~^ 01 = 04 02~^ 



Now 03~^ 01 and 04 02~^ are equiangular motions of opposite type. 

 Since they are equal they must leave all self-complementary complexes 

 invariant and also all anti-self-complementary ones. They then 

 leave all complexes invariant and so 



03~^ 01 = 04 02~^ = =t= /, 



whence 



01 = =■= 03, 02 = ='= 04- 



We shall call 0i and 02 the components of <^. 

 Suppose 



0= 01 02, 0' = 01 02, 



01 and 0( being equiangular rotations of one type and 02 and 

 of the opposite type. Since 0i and 0iare comm utative with 02 and 



0' = 01 01 •02 02- 



The product of two rotations is thus obtained by multiplying together 

 corresponding components. If and 0' are commutative, either 



01 01 = 01 01, 02 02 = 02 02 



or 



01 01 = — 01 01, 02 02 = — 02 02. 



