56 E. B. Wilson, 
As ® possesses six degrees of freedom, a count of constants would 
indicate that the two conditions could be satisfied. The actual 
computation is conducted as follows.1 
2Q-=—a-aja +ae\e +ayly +aAd|C 
Qs =8a+a—, QF'=38a-!+ 4, 225 = 25 ,— 380° +3a-, 
B—=a(2a+a-—*) a|a+a(2Qa+a-%) | +a (2a+a-%)y|y +3a—-2 60, 
2) —$ Qo sl=4(@ —a-) [ala + ele +yly —36|6] 
2 — 2-14 (Qs, — Q-") 4 (a— a1) (aa)? (ala +816 +717) 
— (a+ 4+ a) 6|d| 
From these expressions, it is found that [2 — 2-1— 4 (2, —@7!) Io 
and [E'?)— 4} 2 s/|o are § (a—a—!) [AxatAyp+Azy— Awd) 
and 4(a?— a-?) [xa+yP+2y—3wd| 
where A—(a+a-—), A =a@+4+a-2, o=xat+ypt+ey+wo. 
Neither of these vectors vanishes identically unless a— a—!1—0 or 
a’? —a~?=—0, that is, unless a is a square root or fourth root of 
unity. From the form of the vectors it is clear that they cannot 
be collinear unless 
VAT eas a+2a-2 a?+4+aq-2 
| fe 18 1 3 
In the first case a must be a sixth root of unity, and in the second 
the two vectors are both parallel to o. As the constant a in 2 may 
be arbitrary, the first case can be excluded, and the second violates 
the condition oo=|=0. Hence it may be assumed that the two 
vectors are independent and determine a plane through which any 
6 must pass. But this plane clearly contains @ inasmuch as any 
three rowed determinant from the matrix 
Ax Ay Az Aw 
2 y ly O20) 
oe Vie ae w 
=2(a+1+a-*) =—0 or w— ©. 


vanishes. Hence again the condition oo =|= 0 is violated, and it is 
evident that despite the six degrees of freedom, no reflection @— 
I—2o|o can be chosen such that the conditions (104) may be 
fulfilled, and Q@® may (possibly) be resoluble with two reflections. 
It is necessary to try a different type of reflection. As a matter of 
fact Smith’s theorem happens to be applicable to this particular case. 
The detailed discussion of the various difficulties which arise in 
the different special cases must be postponed to a later time. There 
is one question which will be suggested and left as an easy exercise 
in the use of the dyadics = It is geometrically apparent that, if 

1 Tt may be noted that in the fourth line down X5 = HO) —} 6 al 
would be the negative of the given value as it should be. 

