Boy's surface is a different embedding of the projective plane, and there is a nice continuous deformation from one to the other. If I'm not mistaken, though, Boy's surface only has one axis of symmetry.
Here's a picture of all three: http://www.geom.umn.edu/zoo/toptype/pplane/ -phma
Boy's surface is an immersion but not an embedding (it has no singularities, but double points and a single triple point), whereas the Roman Surface is not an immersion as it has singularities (points where there is no plane tangent to the surface).--Mark grant 13:44, 17 August 2006 (UTC)
Could somebody please explain/resolve the discrepancies between the parametric equations? The one at the head of the page and the one further down are dissimilar. I also tried them both in Maple v5 and neither produced the image of the Roman surface shown on the page. Many thanks. S.
Some problems with the article
1. One problem is that it is not mentioned that a common name for this surface is Steiner's Roman surface. This is far more common than just Roman surface, so it should be mentioned.
2. A more serious problem is that it is described as an immersion of the projective plane into 3-space, but this is not true. The 6 points at the edge centers of the underlying tetrahedron cannot be locally the image of a surface map that is of maximum rank (= 2) there. (This is even implied toward the end of the article where Whitney umbrellas are discussed.) The word immersion is not tossed around carelessly in mathematics: a mapping of a smooth manifold into another one either is or is not an immersion; this is not. (Steiner's map is an immersion of the projective plane on the complement of the preimages of these 6 points.)
(I now notice that Mark, above, also pointed out that this is not an immersion.)
3. The most serious problem is the following paragraph:
The sphere, before being transformed, is not homeomorphic to the real projective plane, RP2. But the sphere centered at the origin has this property, that if point (x,y,z) belongs to the sphere, then so does the antipodal point (-x,-y,-z) and these two points are different: they lie on opposite sides of the center of the sphere. The transformation T converts both of these antipodal points into the same point, If this were true for only one or small subset of points of the sphere, then these points would just be double points. But since it is true of all points, then it is possible to consider the Roman surface to be homeomorphic to a "sphere modulo antipodes", S2 / (x~-x), i.e. a sphere whose antipodal points are equivalent. The real projective plane is known to be homeomorphic to a sphere modulo antipodes, therefore the Roman surface is homeomorphic to RP2.
The fact that antipodal points map to the same point is only part of the reasoning needed to show that the Roman surface is in some sense a projective plane. (This can be seen by the fact that this same argument quoted above works just as well for the map f: S2 → R3 given by f(x,y,z) = (0,0,0) for all (x,y,z) with x2 + y2 + z2 = 1.) The reasoning shows only that the Steiner surface is the image of a projective plane.Daqu 13:57, 21 November 2006 (UTC)
- 1 good point, the lead should be rewritten to reflect this. 2 indeed, I don't know how thats slipped through, is there a good way to describe what it is? 3 indeed, you can be stronger than just an image and you can get a 1 to 1 map from RP2 to the Roman Surface. I've no time now, faancy being WP:BOLD and fixing things. --Salix alba (talk) 17:38, 21 November 2006 (UTC)
- I agree with 1 -- I have seen the surface most commonly referred to as "Steiner's Roman Surface". No clue as to how one would go about changing the name of the article though? I have also tried to clarify that the surface is not an immersion without the removal of the pinch points. If anyone knows a better way to phrase it please make the change!
Incidentally, I also think that Steiner surface should be merged into this article.--A13ean 21:01, 22 July 2007 (UTC)
Another cheerful fact about Steiner's Roman surface
If the 6 singular points of the surface each has a small neighborhood removed, this will expose 6 figure-eight boundaries. If these are grouped into 3 pairs and the pairs piped together (by 3 pipes with figure-eight cross-sections), the resulting surface in 3-space is a bona fide immersion . . . of the non-orientable surface M whose Euler characteristic χ(M) is 6 less than that of the projective plane. That is, χ(M) = 1 - 6 = -5. Since -5 = 2 - 7, this identifies M as the unique non-orientable surface of non-orientable genus = 7 (i.e., M is the connected sum of 7 projective planes). It's not hard to see that this can be done so that M has D3 dihedral symmetry (i.e., the sixfold symmetry of an equilateral triangle). Perhaps this should be mentioned in the article.Daqu 14:38, 21 November 2006 (UTC)
- This is shown for a version of Apery's immersion on the MathWorld page for Boy's Surface. It is attributed to "Wang" but I'm fairly certain someone else did this before -- I'll have to dig up the citation. A13ean 21:01, 22 July 2007 (UTC)