658 - MOORE. 



(12) / = r-[mi(k2ki - hk-z) + m^ikiks - k^ki)] = /■•* 

 where $ = viiikoki — kiki) + mo{kikz — kzki). 



If mi 7^ =>=m2 the only 1-vectors left invariant^ are 



A'l ± iko, ks ± ■/A'4. 



For if r = Xi^'i + X2A'2 + Xs^'s + ^Jc^, 



r' = r-^ = — ??hXiA-2 + 7niK2ki — vioXski + m^kiks 



and if this is to be equal to ixr 



(13) ^niXo = M^i,— mi\i = M^2, ^'^2X4 = M^3,— ^"2X3 = M^4 

 which can be satisfied only for the values 



X2 = ±iXi, X3 = 0, X4 = 0; Xi = Xo = 0, X4 = ±1X3 



If mi = mo we see that (13) is satisfied by any vector of the pencils 



A-i + ik. + X(A-3 + iki), ki - iko + \{kz - iki), 



that is these vectors are left invariant for all values of X. If mi= — nio, 

 the pencils 



ki + iko + X3(A-3 - iks), ki - iko + X(A-3 + iki) 



are left invariant for all values of X. 



If we apply the transformation twice, that is 



/ = r-$, r" = r'-$ = (r-#)-$ 



assuming the above form for r we have 



r" = -Wi-(XlA-i + X2A-2) - 7«2'(X3A-3 + X4A-4). 



If 7»i = ±7??2 r" is a multiple of r and if ??u = ='=???2 = 1, the trans- 

 formation repeated twice is a reflection through the origin and of 

 course repeated four times is the identical transformation. These 

 are the only conditions under which the transformation will close. 

 Hence the necessary and sufficient condition that (12) be of finite order 

 is Ml = ± vio or M* = ± :V. In this case there are two pencils of 

 invariant vectors while in the general case only four vectors are left 

 invariant. 



In order to find the 2-vectors left invariant by the transformation 



9 Invariant here means that r' = lur. 



