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 Roman surface ： ウィキペディア英語版
Roman surface

The Roman surface or Steiner surface (so called because Jakob Steiner was in Rome when he thought of it) is a self-intersecting mapping of the real projective plane into three-dimensional space, with an unusually high degree of symmetry. This mapping is not an immersion of the projective plane; however, the figure resulting from removing six singular points is one.
The simplest construction is as the image of a sphere centered at the origin under the map ''f''(''x'',''y'',''z'') = (''yz'',''xz'',''xy''). This gives an implicit formula of

:$x^2 y^2 + y^2 z^2 + z^2 x^2 - r^2 x y z = 0. \,$
Also, taking a parametrization of the sphere in terms of longitude (θ) and latitude (φ), gives parametric equations for the Roman surface as follows:
:''x'' = ''r''2 cos θ cos φ sin φ
:''y'' = ''r''2 sin θ cos φ sin φ
:''z'' = ''r''2 cos θ sin θ cos2 φ.
The origin is a triple point, and each of the ''xy''-, ''yz''-, and ''xz''-planes are tangential to the surface there. The other places of self-intersection are double points, defining segments along each coordinate axis which terminate in six pinch points. The entire surface has tetrahedral symmetry. It is a particular type (called type 1) of Steiner surface, that is, a 3-dimensional linear projection of the Veronese surface.
==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
:$x^2 + y^2 + z^2 = 1,\,$
we apply to these points the transformation ''T'' defined by $T\left(x, y, z\right) = \left(y z, z x, x y\right) = \left(U,V,W\right),\,$ say.
But then we have
:
\begin
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) \\()
& = (1)(x^2 y^2 z^2) = (xy) (yz) (zx) = U V W,
\end

and so $U^2 V^2 + V^2 W^2 + W^2 U^2 - U V W = 0\,$ as desired.
Conversely, suppose we are given (''U'', ''V'', ''W'') satisfying
(
*) $U^2 V^2 + V^2 W^2 + W^2 U^2 - U V W = 0.\,$
We prove that there exists (''x'',''y'',''z'') such that
(
*
*) $x^2 + y^2 + z^2 = 1,\,$
for which $U = x y, V = y z, W = z x,\,$
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
:$x = \sqrt\right\},\ y = \sqrt\right\},\ z = \sqrt\right\}.\,$
(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 $U^2 V^2 = 0\,$
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
(
*
*
*)$U \neq 0, V = W = 0.\,$
It follows that $z = 0,\,$
(since $z \neq 0,\,$ implies that $x = y = 0,\,$ and hence $U = 0,\,$ contradicting (
*
*
*).)
a. In the subcase where
:$|U| \leq \frac,$
if we determine ''x'' and ''y'' by
:$x^2 = \frac$
and $y^2 = \frac,$
this ensures that (
*) holds. It is easy to verify that $x^2 y^2 = U^2,\,$
and hence choosing the signs of ''x'' and ''y'' appropriately will guarantee $x y = U.\,$
Since also $y z = 0 = V\textz x = 0 = W,\,$
this shows that this subcase leads to the desired converse.
b. In this remaining subcase of the case 3., we have $|U| > \frac.$
Since $x^2 + y^2 = 1,\,$
it is easy to check that $xy \leq \frac,$
and thus in this case, where $|U| >1/2,\ V = W = 0,$
there is no (''x'', ''y'', ''z'') satisfying $U = xy,\ V = yz,\ W =zx.$
Hence the solutions (''U'', 0, 0) of the equation (
*) with $|U| > \frac12$
and likewise, (0, ''V'', 0) with $|V| > \frac12$
and (0, 0, ''W'') with $|W| > \frac12$
(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 $\left(xy, yz, zx\right) = \left(0, 0, 0\right) = \left(U, V, W\right)\,$ as desired.
This covers all possible cases.

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