Page:BatemanConformal.djvu/4

First, let V be a homogeneous function of degree zero. We evidently have

${\begin{array}{lr}{\frac {\partial V}{\partial x}}=&{\frac {\partial V}{\partial l}}+{\frac {\partial V}{\partial \lambda }}+2x{\frac {\partial V}{\partial n}},\\\\{\frac {\partial V}{\partial y}}=&i{\frac {\partial V}{\partial l}}-i{\frac {\partial V}{\partial \lambda }}+2y{\frac {\partial V}{\partial n}},\\\\{\frac {\partial V}{\partial z}}=&{\frac {\partial V}{\partial m}}+{\frac {\partial V}{\partial \mu }}+2z{\frac {\partial V}{\partial n}},\\\\{\frac {\partial V}{\partial w}}=&i{\frac {\partial V}{\partial m}}-i{\frac {\partial V}{\partial \mu }}+2w{\frac {\partial V}{\partial n}},\end{array}}$

${\begin{array}{rl}\left({\frac {\partial V}{\partial x}}\right)^{2}+\left({\frac {\partial V}{\partial y}}\right)^{2}+\left({\frac {\partial V}{\partial z}}\right)^{2}+\left({\frac {\partial V}{\partial w}}\right)^{2}\\\\=&4{\frac {\partial V}{\partial l}}{\frac {\partial V}{\partial \lambda }}+4{\frac {\partial V}{\partial m}}{\frac {\partial V}{\partial \mu }}+4n\left({\frac {\partial V}{\partial n}}\right)^{2}+4l{\frac {\partial V}{\partial l}}{\frac {\partial V}{\partial n}}\\\\&+4\lambda {\frac {\partial V}{\partial \lambda }}{\frac {\partial V}{\partial n}}+4\mu {\frac {\partial V}{\partial \mu }}{\frac {\partial V}{\partial n}}+4m{\frac {\partial V}{\partial m}}{\frac {\partial V}{\partial n}}.\end{array}}$

But, since V is a homogeneous function of degree zero,

$l{\frac {\partial V}{\partial l}}+\lambda {\frac {\partial V}{\partial \lambda }}+m{\frac {\partial V}{\partial m}}+\mu {\frac {\partial V}{\partial \mu }}+n{\frac {\partial V}{\partial n}}+\nu {\frac {\partial V}{\partial \nu }}=0$

and ν = -1, therefore

$\left({\frac {\partial V}{\partial x}}\right)^{2}+\left({\frac {\partial V}{\partial y}}\right)^{2}+\left({\frac {\partial V}{\partial z}}\right)^{2}+\left({\frac {\partial V}{\partial w}}\right)^{2}\equiv 4\left[{\frac {\partial V}{\partial l}}{\frac {\partial V}{\partial \lambda }}+{\frac {\partial V}{\partial m}}{\frac {\partial V}{\partial \mu }}+{\frac {\partial V}{\partial n}}{\frac {\partial V}{\partial \nu }}\right].$

If, instead of (l, m, n, λ, μ, ν), we use the usual hexaspherical coordinates defined by the relations

$a_{1}=x,\ a_{2}=y,\ a_{3}=z,\ a_{4}=w,\ a_{5}={\frac {r^{2}-1}{2}},\ a_{6}={\frac {r^{2}+1}{2i}}$; ${\begin{array}{ccc}l=a_{1}+ia_{2},&m=a_{3}+ia_{4},&n=a_{5}+ia_{6},\\\\\lambda =a_{1}-ia_{2},&\mu =a_{3}-ia_{4},&\nu =a_{5}-ia_{6},\end{array}}$

the equation takes the more symmetrical form

$\left({\frac {\partial V}{\partial x}}\right)^{2}+\left({\frac {\partial V}{\partial y}}\right)^{2}+\left({\frac {\partial V}{\partial z}}\right)^{2}+\left({\frac {\partial V}{\partial w}}\right)^{2}\equiv \sum \limits _{k=1}^{6}\left({\frac {\partial V}{\partial a_{k}}}\right)^{2}.$

This relation shows that a homogeneous function of degree zero, which is a solution of

${\frac {\partial V}{\partial l}}{\frac {\partial V}{\partial \lambda }}+{\frac {\partial V}{\partial m}}{\frac {\partial V}{\partial \mu }}+{\frac {\partial V}{\partial n}}{\frac {\partial V}{\partial \nu }}=0,$

i.e., of

$\sum \limits _{k=1}^{6}\left({\frac {\partial V}{\partial a_{k}}}\right)^{2}=0$