( 493 ) 
So this must correspond with (#6) aq2= 0, which indicates 
according to the above the same pair of rays. By applying the identity 
(a B) (a x) = aa be — ba ay, we find (ab)?bz(a2)=(a b) aa b° — (ab) ba ar bz, 
where the third covariant disappears identically. 
10. The biquadratic form 
at = ayy et + 4 ag) 23 ry + 6 agg af a + 4 aj 2) at agg zi 
has, beside the above mentioned invariant 
h= 0d? an + 2 dog + a4 
and the well known invariants 
t=(ab)* and j=(ab)* (ac)? 0)’, 
the quadratic invariants 
m = at =a*,+ 4 a2 ae Oras 4 ad, + @, 
[== (ab) ae == 2 agg (aag — 2 agg + Aga) — 2 (431 — 443)” « 
In consequence of the identity (ab)? + ay = da by we have 
(a b)* + 2 (aby? a? + at = a2 br, 
a b 
so that we have also 
HAL meh. 
If we put a4, = 0 and a,4 = 0, then 
h = 2 A995 
a 2 (3 ae oa 4 a3) d3) 
j == 6 (2 4g) aag 413459) 
SO 
8A3>=6ith-+ 87. 
