422 MOORE AND PHILLIPS. 



Then let 



Pi = VXi^i + V — \ iq4 



Pi = VXigi — V — X 4g4 



P2 = VXoQ'a + V — X sgs 



Pb = VX2g2 — V — X sgs 



2?3 = VXsgs + V — X ege 



JJe = VX393 — V — Xe^e 



Using these values it is readily seen that 



rs = |(?'iP4 + ^42^0 + h{p2P5 + P0P2) + KpsPs 4- 2J62>3). 



If none of the quantities qi, q-2,. . . , ge are zero this has the form required. 

 If pi is zero we can replace it by any complex and let X be zero in the 

 expression 



rs = \(j?ipi + pipi) + 



Hence every self-conjugate two-two dyadic can be reduced to this 

 form, pu Pi,. . .jJe being complexes. 



Every self-conjugate iico-two dyadic can he expressed as the product 

 of a dyadic and its conjugate. That is, if ^ is any such dyadic, a dyadic 

 ^ = rs can be found such that 



To show this reduce "^ to the form 



^ = \{p)ipi + PiP\) + m(P2P5 + PbP'l) + v{pzPQ + psPs). (A) 



Let qi, q2. ■ .qe be the edges of a tetrahedron, qi and q^, 92 and q^, 

 qz and q^, being the pairs of non-intersecting edges. Choose the mag- 

 nitudes of the g's such that 



[gig-i] = [^295) = [gs^e] = i. 



Now let 



$ = ixiqipi + iU2?2p2 + IJ-Gq^Pe- 



Then 



4>c$ = fXiiJLiipiPi + pipi) + lJ-2lJib(j>2Pb + PbP2) + t^sfJ^eipzPe + P6P3)' 



Comparison of this with (A) shows that we can make $c$ = ■^^ by 

 choosing nx, 1x2. . .^^6 such that 



)UiM4 = X, /XoyUo = /i, A*3M6 = V. 



By using the theorem just proved we can show that the complex g 

 of lines which are transformed into lines is a general quadratic complex, 



