G. PLARR ON THE SOLUTION OF THE EQUATION Vp¢p=0. 53 
2 2 
Sx5=Sy,8—3P8*. 
Now the expression (9) of P may be conceived under the form 
2P=— 2SadSaz’a , 
because we have by =a+6B+t+y: 
and by the definition of y, we have 
Saw’a=—San,, &c. Thus 
2P= + 2SadSayn, = —Sy6 , 
and the first term of Sxé, namely 
S76 = = 
Then as to 6’, we have S'=a? +3’ +y=—3. "his gives : 
Sxk6=—2P+2P=0. 
There remains for 7; : 
Peer a . 
Substituting this into the expression (17) we get : 
—2P?= e—sP* +2Soi7, , 
from which we draw, 
(23) —2P? =3(k? + 2Soy7,) . 
Likewise, in nnPating (21) by Sy(_ ), we get : 
Sym =Snk, 
because 58 vanishes, being nothing but: —2SadSan= + SSawa=Saw’a , 
and this latter result is equal to zero by (8). 
We have thus: 
(25) 3Q=S(ye+ orn). 
Substituting the ere (15), (23), (25), into (20), divided by 3, we get: 
(26) =F = (n° + 2Sor) (kK? + 2S0y7;) —S'(nk + oT +710) . 
§ 4. If there was equality respectively between 7 and a, and between 7, and 
o,, the decomposition of I into a sum of squares would be possible at once. 
We therefore elicit the part of T responding to that hypothesis, in introducing 
into the piace of o, 7, o, 7), the vectors defined by: 
(27) { o=C+e, m=Gt+ea 
7=C—e, ™=G—«. 
VOL. XXVIII. PART I. 0 
