Editing Roman surface (section)

==Derivation of implicit formula==
For simplicity we consider only the case ''r'' = 1. Given the sphere defined by the points (''x'', ''y'', ''z'') such that
:<math>x^2 + y^2 + z^2 = 1,\,</math>

we apply to these points the transformation ''T'' defined by <math> T(x, y, z) = (y z, z x, x y) = (U,V,W),\, </math> say.

But then we have
:<math>
\begin{align}
U^2 V^2 + V^2 W^2 + W^2 U^2 & =  z^2 x^2 y^4 + x^2 y^2 z^4 + y^2 z^2 x^4  =  (x^2 + y^2 + z^2)(x^2 y^2 z^2) \\[8pt]
& =  (1)(x^2 y^2 z^2) = (xy) (yz) (zx) = U V W,
\end{align}
</math>

and so <math>U^2 V^2 + V^2 W^2 + W^2 U^2 - U V W = 0\,</math> as desired.

'''Conversely''', suppose we are given (''U'', ''V'', ''W'') satisfying

(*) <math>U^2 V^2 + V^2 W^2 + W^2 U^2 - U V W = 0.\,</math>

We prove that there exists (''x'',''y'',''z'') such that

(**) <math>x^2 + y^2 + z^2 = 1,\,</math>

for which <math>U = x y,  V = y z,  W = z x,\,</math>

with one exception:   In case 3.b. below, we  show this cannot be proved.

'''1.''' In the case where none of ''U'', ''V'', ''W'' is 0, we can set
:<math>x = \sqrt{\frac{WU}{V}},\  y = \sqrt{\frac{UV}{W}},\  z = \sqrt{\frac{VW}{U}}.\,</math>

(Note that (*) guarantees that either all three of U, V, W are positive, or else exactly two are negative. So these square roots are of positive numbers.)

It is easy to use (*) to confirm that (**) holds for ''x'', ''y'', ''z'' defined this way.

'''2.''' Suppose that ''W'' is 0. From (*) this implies <math>U^2 V^2 = 0\,</math>

and hence at least one of ''U'', ''V'' must be 0 also. This shows that is it impossible for exactly one of ''U'', ''V'', ''W'' to be 0.

'''3.''' Suppose that exactly two of ''U'', ''V'', ''W'' are 0. [[Without loss of generality]] we assume

(***)<math> U \neq 0,  V = W = 0.\,</math>

It follows that <math>z = 0,\,</math>

(since <math>z \neq 0,\,</math> implies that <math>x = y = 0,\,</math> and hence <math>U = 0,\,</math> contradicting (***).)

'''a.''' In the subcase where
:<math>|U| \leq \frac{1}{2},</math>

if we determine ''x'' and ''y'' by
:<math>x^2 = \frac{1 + \sqrt{1 - 4 U^2}}{2}</math>

and <math>y^2 = \frac{1 - \sqrt{1 - 4 U^2}}{2},</math>

this ensures that (*) holds. It is easy to verify that <math>x^2 y^2 = U^2,\,</math>

and hence choosing the signs of ''x'' and ''y'' appropriately will guarantee <math> x y = U.\,</math>

Since also <math>y z = 0 = V\text{ and }z x = 0 = W,\,</math>

this shows that '''this subcase''' leads to the desired converse.

'''b.''' In this remaining subcase of the case '''3.''', we have <math>|U| > \frac{1}{2}.</math>

Since <math>x^2 + y^2 = 1,\,</math>

it is easy to check that <math>xy \leq \frac{1}{2},</math>

and thus in this case, where <math>|U| >1/2,\   V = W = 0,</math>

there is '''no''' (''x'', ''y'', ''z'') satisfying <math> U = xy,\  V = yz,\  W =zx.</math>

Hence the solutions (''U'', 0, 0) of the equation (*) with  <math>|U| > \frac12</math>

and likewise, (0, ''V'', 0) with  <math>|V| > \frac12</math>

and (0, 0, ''W'') with <math>|W| > \frac12</math>

(each of which is a noncompact portion of a coordinate axis, in two pieces) '''do not correspond to any point on the Roman surface'''.

'''4.''' If (''U'', ''V'', ''W'') is the point (0, 0, 0), then if any two of ''x'', ''y'', ''z'' are zero and the third one has [[absolute value]] 1, clearly <math>(xy, yz, zx) =  (0, 0, 0) = (U, V, W)\,</math> as desired.

This covers all possible cases.