Figure and Table Captions

Gammap 1 \Gamma m 1 +m 2 ; M (\Gammap 2 \Gamma m 1 +m 2 ) ; M 4;j = \Gammam 2 ! j for j = 3; 4; 5 ; 5;2 = 1 ; M 5;j = \Gamma! j for j = 6; 7; 8 ; (C24) M 6;j = 1 and M 7;j = ! j for j = 6; 7; 8 ; (C25) 8;1 = \Gammae 8;2 = \Gamma1 : (C26) The solution to Z = S 49 So, when the first two components of the vector identity (C4) is written long hand, we get L 3 [fl j ]X j + ffi l;1 ; (C7) fl j L 3 [fl j ]X j k + 1)X j + ffi l;2 ; (C8) where ffi k;j is the usual Kronecker delta symbol. From the (k + 3)-rd element of the vector identity, with k ranging from 0 to 5, it follows that flD j + ffi k;l\Gamma3 (C9) Using this in (C7) and (C9), we get p j L 3 [p j ]X r l\Gamma3 + ffi l;1 ; (C10) j L 3 [p j ]X r l\Gamma4 + ffi l;2 : (C11) from which it follows that 1 = Now, let's consider the other elements H k;l for k 3. Clearly, from