Talk:Boy's surface

Boy's surface received a peer review by Wikipedia editors, which is now archived. It may contain ideas you can use to improve this article.

Old comments on parameterization

I'd like to propose moving the section on the Structure of Boy's Surface in front of the section on parameterization. Wikipedia is addressed to a general audiance, who are more likely to relate to the pictures than the equations. --agr 15:47, 6 Jun 2005 (UTC)

There being no objection, I went ahead and made the change. --agr 18:01, 21 October 2005 (UTC)[reply]

I think the images could use more transparency and smoothness. Drawing them opaque and polygonized like this makes it harder to see what's happening. There's a sketchy idea of what I mean on my page [1] but maybe a raytrace would look better. I'd also like to see a more prominent treatment of the locus of self-intersection points (a single triple point connecting three loops of double points). —David Eppstein 21:56, 13 September 2006 (UTC)[reply]

I put together: [2] this afternoon. If you think that looks right feel free to insert it as you see fit (I assume you have done most of the organizing of this article.) Let me know if you would like a smaller verision too, it does seem rather big now. — A13ean (talk)

Pretty! And a lot more understandable than what's there now. I'll work on getting it in and making the other images a little less space-consuming. —David Eppstein 05:03, 4 November 2006 (UTC)[reply]

Seems like Boys surface is used as a halfway stage in a Sphere eversion http://torus.math.uiuc.edu/jms/Papers/isama/color/opt2.htm, something to this effect should be included. --Salix alba (talk) 22:45, 13 September 2006 (UTC)[reply]

Doing some google digging finds [Imaging maths - Unfolding polyhedra] which talks about a discreet Boy's surface, with a reference

U. Brehm, How to build minimal polyhedral models of the Boy surface. Math. Intelligencer 12(4):51-56 (1990).

Bryant / Kusner

It should be noted that the article by Hermann Karcher and Ulrich Pinkall which can be found here attributes the parametrization formula to Rob Kusner instead of Robert Bryant. Instead it credits Bryant with a general result about rational immersions of some kind of minimal surfaces, of which this parametrization is a special case apparently. regards, High on a tree 21:05, 6 December 2006 (UTC)[reply]

It looks like the parameterization used for the statue is from Kusner, which is (as far as I can tell) different from the parameterization currently given in the article which is attributed to Bryant. That parameterization with the credit to Bryant also appears in Apery's book. If there are no objections, we can change the mention in the "statue" section. Thanks, High on a tree! Sorry this is rather late. A13ean 02:35, 26 June 2007 (UTC)[reply]

Upon even more in depth digging it turns out that this class of parameterizations is due to a collaboration between Bryant and Kusner, and they are commonly refered to as "Bryant-Kusner parameterizations". So a compromise it is, and unless no one objects I'll make the change. A13ean 02:38, 20 September 2007 (UTC)[reply]

Javaview model

The Boy Surface

Here's a model of Boy's that one can move around, for those of us who like to play with things. If anyone else thinks that it's worth it feel free to link it, but I'll refrain since I made it. — A13ean (talk)

I like it! It's much more helpful IMHO to have something one can rotate freely instead of seeing it animated. So I'll add it. --C S (Talk) 23:36, 13 May 2007 (UTC)[reply]

Left and right handed boys

Something should be said about left and right-handed Boy's surfaces, which are not regularly homotopic. --C S (Talk) 11:17, 8 April 2007 (UTC)[reply]

False claim

The page says that the Boy surface is homeomorphic to RP^2, but that is just impossible, as there are a compact non orientable surface cannot be embedded in R^3... —Preceding unsigned comment added by 201.252.9.131 (talk • contribs)

It's not embedded, it's immersed. —David Eppstein 00:03, 19 July 2007 (UTC)[reply]

That's irrelevant. The problem is with the word homeomorphic, which is evidently wrong.

In fact, the argument presented in the "Relating the Boy's surface..." section is bogus: what's said there does not imply that the parametrization is injective in the interior of the disk. Mariano (talk) 02:47, 3 December 2008 (UTC)[reply]

Hilbert

I changed 'Hilbert' in the article to 'David Hilbert', hoping that it would help those who (like myself) weren't aware of who 'Hilbert' referred to. I am not, however, certain that it is indeed in reference to David Hilbert, but his biographical information (lifetime, profession, nationality, etc.) strongly suggests that this is true. Please correct my edit if I am mistaken. Thank you. Spinnick597 00:38, 28 September 2007 (UTC)[reply]

That's the right one. A13ean 11:05, 28 September 2007 (UTC)[reply]

Parametrization and implicit equation

The given parametrization of Boy's surface is either inconsistent or some essential information is lacking: the coordinates of a point are expressed rationally in terms of $g_{1},g_{2},g_{3},$ which are themselves irrational functions of the real and imaginary parts of a complex number, named by the same letter as the third coordinate of the points of the surface. The problem is that, the expression of $x,y,z$ in terms of $g_{1},g_{2},g_{3}$ implies immediately that $x^{2}+y^{2}+z^{2}=1.$ Thus this complicate parametrization is simply a strange parametrization of the unit sphere.

No, because the denominator is

g_{1}^{2}+g_{2}^{2}+g_{3}^{2}

not

{\sqrt {g_{1}^{2}+g_{2}^{2}+g_{3}^{2}}}

. —Tamfang (talk) 07:38, 8 June 2014 (UTC)[reply]

By the way, beside a lacking correct parametrization, other fundamental properties are lacking: I believe that Boy's surface is algebraic. Therefore, an implicit equation must be provided, or, at least, its degree must be given (from the picture, I guess 6). Also, it must been said if Boy's surface admits a rational parametrization. I guess that it is true, because, otherwise, Boy's surface would not be isomorphic (that is birationally equivalent) to the projective plane. D.Lazard (talk) 15:35, 4 April 2014 (UTC)[reply]

I'm also bothered that the Bryant parametrization has a boundary. Could someone explain how that is? —Tamfang (talk) 07:38, 8 June 2014 (UTC)[reply]

This is explained in section "Property of Bryant-Kusner parametrization": this is a two fold parameterization, and the condition that the magnitude of z is not greater than 1 is here only for having a bijective parameterization (outside the unit sphere). D.Lazard (talk) 09:44, 8 June 2014 (UTC)[reply]

I have added an external link to a page that answers to my preceding questions. It remains to edit the article to include these answers. D.Lazard (talk) 09:44, 8 June 2014 (UTC)[reply]

two-period parametrization

I wish I had the skill to derive a parametrization based on trig functions and the stitching illustrated here. Has no one done so? Would it be a double or a quadruple cover? —Tamfang (talk) 05:14, 29 June 2014 (UTC)[reply]

Given the three-fold symmetry of the usual embeddings, wouldn't it make more sense to start with a hexagon with opposite edges glued, instead of a square? —David Eppstein (talk) 05:17, 29 June 2014 (UTC)[reply]

The first external link of the article contains various parameterizations and references to the articles where they are defined. It explains also (in French) why Boy's surface differs from other realizations of projective plane (all singular points are simple crossing points — no cusps). I guess that the answer to your question is in this link and its references. By the way, it would be useful to expand the article, using this link. D.Lazard (talk) 09:27, 29 June 2014 (UTC)[reply]

IMHO the easiest way to derive a parametrization is to start with the unit circle, and write x(theta, phi), y(theta, phi), z(theta, phi) making sure that each coordinate function is periodic in phi -> pi - phi. This last part is pretty easy, you can chose a subset of the spherical harmonics that satisfy it. This is how John Hughes found his parametrization of Boy's surface -- he projected a 7-vertex polygon model on the spherical harmonics to solve for the three functions, also taking use of the three-fold symmetry. This is also how the parametrization for the other immersion of the real projective plane was found, although in that case there's not any three-fold symmetry. a13ean (talk) 17:32, 1 July 2014 (UTC)[reply]

Eliminating confusion caused by a poor choice of variable names

Given that x, y and z are conventionally used as the three spatial coordinates in Cartesian 3-space, it would be wise to choose some letter other than z to represent the complex variable of the definition. Doing so would avoid confusing the non-mathematicians in our audience (and also promote clarity and rigour among the mathematicians!) I suggest using the letter w for the complex variable, as is often done. This would obviously require a thorough edit of the article, at least the Parametrization section, to replace all instances of z (except for those that refer to the third spatial coordinate) with w.

yoyo (talk) 13:22, 29 October 2015 (UTC)[reply]

I agree. Moreover, it is written z(z) in the article. I'll do that. D.Lazard (talk) 14:47, 29 October 2015 (UTC)[reply]

Construction

I find the "Construction" section pretty obscure. Some diagrams seem pretty necessary. Does anyone know where that's from? --Dylan Thurston (talk) 15:20, 4 October 2016 (UTC)[reply]

Connection to Borromean rings

I realised that there is a way to use Borromean rings to help visualise the immersion:-

- First, set up or imagine the rings as symmetrical ovals, e.g. ellipses, each placed symmetrically in a plane orthogonal to the others.

- Then look at them along the body diagonal of a cube symmetrically containing them, to see the threefold rotational symmetry.

- Then bisect them at a plane orthogonal to that diagonal and discard one part.

- From the set of three ovals, each enclosing the next, that gives you three roughly J shaped sloping and slanting arches, with their feet forming a regular hexagon, each arch going over the next and linking opposite vertices of the hexagon.

- Looking at the arches two at a time lets you see that each strip bridging opposite edges of the hexagon can be set up to fit with the others, though the creasing still needs a straightforward rounding off.

Is this merely a visual aid that at least helped me, or is there a deeper association? And is it worth it/acceptable to put any of this in the article? PMLawrence (talk) 04:15, 19 July 2020 (UTC)[reply]

New Video

I created a new video [3], which shows the construction from a Mobius strip, various slices of u and v. Not sure if it is worth it to add a link to the main page, or upload the video. I'd need some advice. Coolwanglu (talk) 20:20, 3 November 2023 (UTC)[reply]