# Talk:Real projective plane

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## Informal construction

Can someone clean this section up and perhaps give it a more thorough treatment? For example, what is a circle of "length" 2 * pi? I'm assuming by length, they mean circumference. Vecter (talk) 20:06, 3 January 2008 (UTC)

it's not clear at all what point that section is trying to make. i am going to remove it. Mct mht (talk) 17:48, 4 January 2008 (UTC)

## Homogeneous coordinates

Here:

The set of lines in the plane can be represented using homogeneous coordinates. A line ax+by+c=0 can be represented as (a:b:c).

whyare we speaking about "lines in the plane"? What has it to do with the projective plane?--Pokipsy76 16:06, 21 July 2006 (UTC)

It's probably in there because the real projective plane can also be thought of as the collection of all one-dimensional subspaces of R^(n+1) —Preceding unsigned comment added by 68.184.213.58 (talk) 06:43, 30 November 2007 (UTC)

## Embedding in R^3

The Generalized Jordan Curve Theorem only applies to (co-dimension 1) images of spheres, not to real projective spaces or quotients of spheres, if you like. Is the idea that an embedding of RP^2 in R^3 can be pulled back to a map from S^2 to R^3? I'm unclear about this. 169.233.53.103 16:36, 3 April 2007 (UTC)

The generalized jordan curve theorem says any compact connected boundaryless n-dimensional manifold, when embedded in R^{n+1} seperates it into two components, one bounded, the other unbounded. In fact, you can replace R^{n+1} with any simply-connected manifold. The proof is essentially homological and appears in most intro alg-top texts. I like the proof in the Guillemin and Pollack differential topology textbook. Here is a rough sketch, assuming everything is smooth: let M be the submanifold of R^{n+1}. First of all, M is closed proper subset of R^{n+1}. Let * be a point of R^{n+1}-M. Let x be another point of R^{n+1}-M. Let p be a path, starting at * and ending at x. Let #(p) be the number of times p intersects M, mod 2. This is well-defined for a generic p by transversality/sards theorem, and it does not depend on the choice of p because R^{n+1} is simply connected -- given a 1-parameter family of functions p_t, you can assume it intersects M transversely and then you notice that the preimage of M under that 1-parameter family is a 1-dimensional submanifold in the parameter domain. FYI, I added this version of the GJCT to the Wikipedia page for JCT. Rybu 16:02, 8 July 2007 (UTC)
Oh, I forgot to add one thing: now conzider this Z/2-valued invariant as a function of the point x. Then it is locally constant, takes value 0 near *, and also takes the value 1 at some points (consider it in a neighbourhood of a point of M). That's the proof. Rybu 16:02, 8 July 2007 (UTC)
All very well, but the piped link Generalized Jordan Curve Theorem doesn't lead to the required information. Charles Matthews (talk) 10:03, 29 July 2009 (UTC)
Your link is wrong -- the one in the article is: Generalized Jordan Curve Theorem. Rybu (talk) 17:26, 4 August 2009 (UTC)

## Formal Construction

"Consider a sphere, and let the great circles of the sphere be "lines", and let pairs of antipodal points be "points". [...] This is the real projective plane." The omitted portion is not relevant, other than as an argument for the validity of the first statement. As someone not familiar with topology, but who understands what both great circles and antipodal points are, I have no idea how to use great circles as "lines", or how this makes a plane. --68.76.222.122 03:21, 29 April 2007 (UTC)

## Möbius strip?!

I don't get how this is a Möbius strip. The Möbius strip is only attached in one direction. It's an OPEN surface. The Klein bottle is connected on both direction, like the square and arrow diagram shows! Tom Ruen 05:36, 21 July 2007 (UTC)

Okay, I see a Klein bottle is only inverted on one axis and closed in the other, just like the Möbius strip is inverted in one direction but OPEN in the other. So neither exactly represents the projective plane. I guess the Möbius strip is a better representation giving half the connection and imagining the other applied to the open edge. Tom Ruen 18:01, 22 July 2007 (UTC)
(edit conflict) What that paragraph is trying to say is that if you glue the edge of the Möbius strip to itself, thereby closing it, in the right way, you get a projective plane — or at least you would, if you could do that without having to worry about those pesky self-intersections you get in 3D. I've tried to edit the paragraph to make the intended meaning clearer. Any further improvements would be most welcome. —Ilmari Karonen (talk) 18:10, 22 July 2007 (UTC)
Thanks. I got it, just read too quick sometimes. Better wording now - no perfect tricks to make dumb people like me read slower. I'm just wondering about the "fundamental polygon" image. I see ANY even-sided regular polygon can be fundamental, with opposite sides corresponding, or in the limit a circle with center-point reflection defining the correspondence. Unit circle [r,θ]: [1,θ]=[1,θ+180]. Maybe I just like pictures, but I think that would be helpful to see as well. Tom Ruen 18:39, 22 July 2007 (UTC)
Yes, certainly. I originally drew those squares as a series for the fundamental polygon article; as it happens, only one of them is currently used there, and another one seems to have been replaced with a slightly different version. I made them all squares since a square is sufficient to construct both the sphere, the projective plane, the torus and the Klein bottle, and I wanted the constructions to be as directly comparable as possible.
As it happens, the minimal fundamental polygon for the projective plane, in an algebraic sense, is two-sided — but of course a meaningful drawing of such a degenerate polygon is impossible without at least cheating a bit. There is another series of fundamental polygons on Commons that does show two-side polygons — with curved sides — for the sphere and the projective plane.
I do agree that nicer illustrations could probably be drawn for this article. In fact, you've given me some ideas above: perhaps a circular disc with two arrows going around it, or maybe (also?) even a real Möbius strip with similar arrows. In the mean time, if you can think of better illustrations, please do draw some or, if you don't wish to draw them yourself, suggest them here. You could also consider asking at Wikipedia:WikiProject Mathematics/Graphics; it's not a very busy place, but there are some excellent illustrators hanging around there. —Ilmari Karonen (talk) 19:44, 22 July 2007 (UTC)
I can produce images of all four immersions of the real projective plane in R^3 with a single triple point, if that would be useful. I'll post them here to see if they can be worked into the article. Actually, there's not a mention of the fact that there are exactly four ways to immerse it in R^3 (again with a single triple point). Maybe I'll add a section if anyone else thinks it deserves a mention. Apery has a proof, but I pity anyone who actually has to read it. -- A13ean 21:17, 22 July 2007 (UTC)

I suggest moving the 2nd paragraph on the Moebius strip based construction into the "informal construction" section. Perhaps what's in there right now might be combined in. An illustration might help too -- maybe the Sudanese surface with a disk being sewn on to the last edge? Seems like it might be hard to see with the disk being attached to the normal Moebius strip. -- A13ean 21:30, 22 July 2007 (UTC)

## "Embedding" in ${\displaystyle \mathbb {R} ^{4}}$ example

In the "embedding" example (CLICK HERE) there's something I don't understand: it doesn't seem to be an embedding. If you compute the Jacobian matrix you get a 4×3 matrix given by

${\displaystyle J(x,y,z)=\left({\begin{array}{ccc}y&x&0\\z&0&x\\0&2y&-2z\\0&2z&2y\end{array}}\right).}$

Now this matrix doesn't always have rank three, i.e. the "embedding" isn't an immersion. Letting y = z = 0 gives

${\displaystyle J(x,0,0)=\left({\begin{array}{ccc}0&x&0\\0&0&x\\0&0&0\\0&0&0\end{array}}\right),}$

which away from x = 0 has rank two, and when x = 0 has rank zero. From the article on immersions (CLICK HERE) we have "A smooth embedding is an injective immersion f : M → N which is also a topological embedding,..." So clearly the example of an "embedding" isn't an embedding because it's not even an immersion.

Dharma6662000 (talk) 18:57, 25 August 2008 (UTC)

You seem to have lost the point somewhere. y=z=x=0 is not on the projective plane, so you're talking about a map whose domain is not RP^2. Rybu (talk) 21:40, 29 August 2008 (UTC)
No, s/he hasn't lost the point. The matrix has non-maximal rank when y = z = 0. So let y = z = 0 and choose x non-zero. The rank is then two which is less than three, which is non-maximal. Declan Davis (talk) 21:47, 29 August 2008 (UTC)
To be specific, the Jacobian at (1,0,0) (on the projective plane) is degenerate, which seems bad to me, though I don't really get this sort of stuff. JackSchmidt (talk) 22:02, 29 August 2008 (UTC)
It looks like the map is supposed to be from ${\displaystyle \mathbb {RP} ^{2}}$ and not from ${\displaystyle \mathbb {R} ^{2}}$. I propose to change the example so that ${\displaystyle (x:y:z)}$ replaces ${\displaystyle (x,y,z)}$ since the use of colons is more conventional when it comes to projective space. I also propose to mention that we choose affine charts ${\displaystyle (x:y:1)}$, ${\displaystyle (x:1:z)}$, and ${\displaystyle (1:y:z)}$ to give maps from ${\displaystyle \mathbb {R} ^{2}}$ to ${\displaystyle \mathbb {R} ^{4}}$. This would clear things up. Any objections? Declan Davis (talk) 22:28, 29 August 2008 (UTC)
There seems to be a major misperception somewhere. For this map to be an immersion, the derivative has to have maximal rank from the *tangent space* of RP^2 to the tangent space of R^4. RP^2 is two dimensional, so that means that when you restrict the above matrix to the tangent space of RP^2, it must have rank 2. Since the matrix already has rank 2, the only thing to check is that the kernel of the matrix is not-tangential, which is an easy enough exercise in solving simultaneous equations. I refrained from using homogeneous coordinate notation (x:y:z) for a good reason, the map is defined as a 2:1 map out of S^2. If you start by using homogenous coordinates you need to normalize the map, which makes the exposition messier and makes it even harder to see why the map is an embedding. I think you two should really figure out why this map is an embedding before you mess around with the exposition. Rybu (talk) 22:40, 29 August 2008 (UTC)

No-one's messed around with anything. I've simply made a comment on a discussions page. I suggest you read the Wikipedia guidelines for discussions. Be welcoming, assume good faith, etc. You may well be right in what you say, but there is not enough explanation in the article: if there were then people wouldn't be getting confussed. I work in differential geometry, and I am no novice, but the lack of explanation was enough to cause confussion in my mind. The article needs to be changed to improve upon the explantion. Why don't you include what you have just written? But in a more welcoming and friendly tone of course. Also, there's no need to check that the kernel is or is not tangential. The domain of the differential is the tangent space to the initial manifold, and so one simply needs to check that the differential has maximal rank. Since the domain is the tangent space the kernel is always tangential. The kernel may be zero-dimensional (a point in the tangent space to the initial manifold), it may be one-dimensional (a line in the tangent space to the initial manifold), etc. A subspace of the tangent space is always tangent... it's the image we're interested in.  Declan Davis   (talk)  15:06, 21 September 2008 (UTC)
${\displaystyle J(x,y,z)=\left({\begin{array}{ccc}y&x&0\\z&0&x\\0&2y&-2z\\0&2z&2y\end{array}}\right).}$

is the derivative of the map

${\displaystyle (x,y,z)\longmapsto (xy,xz,y^{2}-z^{2},2yz)}$

which is being thought of as a function ${\displaystyle \mathbb {R} ^{3}\to \mathbb {R} ^{4}}$. The induced map out of RP^2 comes from two steps: 1) restricting to the unit sphere ${\displaystyle x^{2}+y^{2}+z^{2}=1}$ and then 2) taking the induced map on RP^2. To check that the induced map out of RP^2 is an immersion one needs to only ensure that map out of ${\displaystyle S^{2}}$ is an immersion (since quotient maps are local diffeos), which means that the kernel of the above matrix (ie: a linear map from R^3 to R^4) does not contain vectors orthogonal to ${\displaystyle (x,y,z)}$ where ${\displaystyle x^{2}+y^{2}+z^{2}=1}$. Anyhow, I made some changes to the article that breaks the description of the map up into simpler steps. Hopefully that'll help. Rybu (talk) 19:11, 21 September 2008 (UTC)

It does seems better now Rybu. Although do you not think that it might be better to use the standard affine charts x ≠ 0, y ≠ 0, and z ≠ 0. It doesn't change the validity of the statement, but it might make it easier to understand. What are your views?  Declan Davis   (talk)  20:12, 21 September 2008 (UTC)
Why don't you show us what you mean here? My immediate impression is that charts will only complicate things. Rybu (talk) 20:33, 21 September 2008 (UTC)

Right, well, ${\displaystyle \mathbb {RP} ^{2}}$ is given by the triple ${\displaystyle (x:y:z)}$ where either x ≠ 0, y ≠ 0, or z ≠ 0. By the definition of projective space we have ${\displaystyle (x:y:z)=(\lambda x:\lambda y:\lambda z)}$ for all non-zero real λ. Assume that x ≠ 0 then we can divide through by x. We have ${\displaystyle (x:y:z)=(1:y/x:z/x).}$ This is one of the so called affine charts. It is a coordinate chart in the usual sense, just like

${\displaystyle (u,v)\mapsto (u,v,{\sqrt {1-u^{2}-v^{2}}})}$

gives a coordinate chart for the two-sphere. So in this x ≠ 0 chart we can replace x with 1, y with ${\displaystyle {\tilde {y}}}$, and z with ${\displaystyle {\tilde {z}}}$. The map in question then becomes

${\displaystyle ({\tilde {y}},{\tilde {z}})\mapsto ({\tilde {y}},{\tilde {z}},{\tilde {y}}^{2}-{\tilde {z}}^{2},2{\tilde {y}}{\tilde {z}}).}$

This is clearly an immersion. Similarily for y ≠ 0 and z ≠ 0. The standard example is that of conics. Take a homogeneous polynomial of degree two in x, y, and z. Depending on the choice of the line at infinity we set x = 1, y = 1, or z = 1. This gives a local chart for projective space. The projective plane is locally just the plane. Once you check it for each of the charts then you know that the whole thing works. There are three affine charts. If you normalise by using the sphere then you need to take six charts, these are

${\displaystyle (u,v)\mapsto (u,v,\pm {\sqrt {1-u^{2}-v^{2}}}),}$
${\displaystyle (u,v)\mapsto (u,\pm {\sqrt {1-u^{2}-v^{2}}},v),}$
${\displaystyle (u,v)\mapsto (\pm {\sqrt {1-u^{2}-v^{2}}},u,v).}$

So to prove the map is an immersion one needs to check six cases all with radicals involved. Taking affine coordinates gives three cases with only linear substitutions.

Declan Davis   (talk)  23:31, 21 September 2008 (UTC)

Thanks. So how does this help the exposition? All you've done is describe charts for S^2 and RP^2, not involving the original map in a non-trivial way, so as far as I can see your efforts are largely beside the point of describing the embedding. Every claim -- that it's an immersion, or that it's 1-1 out of RP^2 -- are easier to verify from the original function. If this section were about charts, I'd understand. But it's not, it's about the map ${\displaystyle (x,y,z)\longmapsto (xy,xz,y^{2}-z^{2},2yz)}$. This map is about as simple as you could ask for -- it's a quadratic polynomial. It's globally defined. For the sake of argument, consider the map from the orthogonal group to itself where you take a matrix ${\displaystyle A}$ and map it to ${\displaystyle A^{2}}$. Would you consider it more informative to use charts for ${\displaystyle O(n)}$ and describe the map in terms of those charts? I'm confused. Rybu (talk) 00:15, 22 September 2008 (UTC)
I like your (DD) idea to use the planes for the local definition. In other words, to check that (x,y,z) -> (xy,xz,yy-zz,2yz) defines an immersion of RP^2 it suffices to check it on the planes (1,y,z), (x,1,z), and (x,y,1), where it is very easy to check. Introducing the sphere makes it easy to complicate things with square roots. I think Rybu is envisioning a method that does not need the square roots, but if one were to use charts on the sphere (as I think many people might try), then things get nasty. Basically, Rybu's method doesn't need charts, but choosing the sphere as the covering space means that your choice of charts is pretty bad. DD's method of using R^3 - 0 as the covering space means that the charts can be chosen very, very nicely, but makes it harder to use Rybu's tangent space method (since the map from R^3-0 to RP^2 loses a dimension).
Here is the check for DD's method: On the plane x=1, the map is (y,z) -> (y,z,yy-zz,2yz) and you can stop reading after the first two coordinates, clearly an immersion. On the plane y=1, the map is (x,z) -> (x,xz,1-zz,2z), and you only need to check the first and last coordinate to see it is an immersion. On the plane z=1, the map is (x,y) -> (xy,x,yy-1,2y) and the second and fourth coordinates suffice to check the immersion. Using these three planes, the map is obviously an immersion since it very nearly the identity on suitable subspaces.
Note that Rybu's method is just a projection of a null space onto a linear subspace. Other than needing to operate over a weird field like R(x,y,z), it is very easy linear algebra. Both methods seem reasonable to me. In this particular case, I prefer DD's method since it uses planes as charts (and the identity matrix as the immersion), but I think it is clear that in general choosing charts gets very messy. JackSchmidt (talk) 02:06, 22 September 2008 (UTC)
Rybu: I mention charts because if you recall Dharma6662000 was confussed by the map, and thought that it was a map from three-space to four-space. If you treat it as such then it's not an immersion. To clear up any chance of confussion I suggest that some mention of charts be made so as it make it crystal clear to the uninitiated reader. The basic way of checking (and showing) that something is an immersion is to cover the domain with charts and then check that the map's an immersion in each of these charts.
Declan Davis   (talk)  13:27, 22 September 2008 (UTC)
Declan Davis   (talk)  21:23, 22 September 2008 (UTC)
We have a problem here where there is no clear way to measure what is simple or for the masses. You are leaning towards a projective geometry viewpoint, which I have repeatedly said is beside the point, at least at the current suggested level of exposition. IMO you'd do well you refrained from inaccurately quoting me. Anger doesn't help. The main point I've been driving at is that your usage of charts requires immense amounts of notation to achieve less than what Dharma6662000 achieved in his post that started this whole discussion. So how does it help? I'm not against a projective geometry approach to all this, but I think if we're going to do something it needs to be reasonably concise and insightful. Rybu (talk) 22:12, 22 September 2008 (UTC)
Common sense should prevail. I don't know about your background, but I've been teaching in a university for some time now and for me simple or for the masses is what I teach my earlier level students, and how I teach it to them. On a second point, I have not misquoted you, and I am not angry, quite the opposite. Would you like me to post some links to your previous contributions which show your grumpy and bad tempered remarks? I have been on the receiving end of them, and I thought it was just me. But I’ve spoken to some other editors and they seem to have the same impression as me. I didn't want to leave any links as not to embarrass you, but if you would like some evidence then I could leave many for the Wikipedia community to see. Please ask if you would like me to so.
I have put together a prototype for a replacement example, any comments?

The real projective plane embeds into four-dimensional Euclidean space. Consider the map ${\displaystyle f:\mathbb {RP} ^{2}\to \mathbb {R} ^{4}}$ given by ${\displaystyle f:(x:y:z)\mapsto (xy,xz,y^{2}-z^{2},2yz)}$. To show that this map is an immersion we simply check that its restiction to the three affine charts given by ${\displaystyle \{(x:y:z):x\neq 0\}}$, ${\displaystyle \{(x:y:z):y\neq 0\}}$, and ${\displaystyle \{(x:y:z):z\neq 0\}}$ are immersions. These charts cover ${\displaystyle \mathbb {RP} ^{2}}$ since either x ≠ 0, y ≠ 0, or z ≠ 0 by definition. This is possible since locally the real projective plane looks like the Euclidean plane. Consider the chart x ≠ 0, then we have ${\displaystyle (x:y:z)\sim \left(1:{\frac {y}{x}}:{\frac {z}{x}}\right)}$. Relabelling ${\displaystyle {\tilde {y}}:=y/x}$ and ${\displaystyle {\tilde {z}}:=z/x}$ we get ${\displaystyle f:(1:{\tilde {y}}:{\tilde {z}})\mapsto ({\tilde {y}},{\tilde {z}},{\tilde {y}}^{2}-{\tilde {z}}^{2},2{\tilde {y}}{\tilde {z}}).}$ This is locally a map from two-space to four-space and can be treated as such. Since the first two coordinates are ${\displaystyle {\tilde {y}}}$ and ${\displaystyle {\tilde {z}}}$ it follows that the map is an immersion in this x ≠ 0 chart. Similar claculation show that f is an immersion in all other affine charts. Notice that this embedding admits a projection into ${\displaystyle R^{3}}$ which is the Roman surface.
Declan Davis   (talk)  22:27, 22 September 2008 (UTC)
One problem with the above is that it's not a well-defined map, since ${\displaystyle (1:1:1)}$ does not get sent to the same place as ${\displaystyle (2:2:2)}$. IMO we're getting at the nature of our misunderstanding by this exercise. Your further chart examples suffer from this problem, too. Rybu (talk) 22:41, 22 September 2008 (UTC)
You're right there: good point! But then the map can't be well defined using the lines-through-the-origin model since by definition (x : y : z) is the same point as (λx : λy : λz) for all non-zero real λ. The article does start of by saying that "In mathematics, the real projective plane is the space of lines in R3 passing through the origin." But it seems that the map in the example is well defined only if you take the two sphere and identify antipodal points. But that in itself is taking a (semi-global) chart. So basically we have that situation that you don't want to use a choice of charts because they're complicated, but the example only works if we take a certain kind of (semi-global) chart to begin with. It's a strange cyclical situation that we have. Personally I don't like using the quotiented two-sphere because it doesn't enable any direct computations for the beginner, but hey, that's my opinion and nothing more.
Declan Davis   (talk)  00:10, 23 September 2008 (UTC)
semi-global chart well, I wouldn't call it that. It's just an equivalent definition of the projective plane. Getting back to the exposition: you can fix your approach by making it more complicated. Define the map by ${\displaystyle (x,y,z)\longmapsto ({\frac {xy}{x^{2}+y^{2}+z^{2}}},{\frac {xz}{x^{2}+y^{2}+z^{2}}},{\frac {y^{2}-z^{2}}{x^{2}+y^{2}+z^{2}}},{\frac {2yz}{x^{2}+y^{2}+z^{2}}})}$. Now you can get back to checking its an immersion, 1-1, etc. This is what I was getting at -- by forcing the exposition into this language you create more pitfalls for the reader. Maybe it would make sense to include both "versions" of the embedding, so that people are free to think of it either way? Rybu (talk) 03:56, 23 September 2008 (UTC)
Hmmm, I see what you mean. It's not the easiest thing to work with. But this does also show that the sphere model is taking a semi-global chart. If we use ${\displaystyle \mathbb {S} ^{2}/\{\pm 1\}}$ to parametrise all of the direction through the origin (i.e. chose a chart) then this map reduces to the old one. I guess the example's better left as it is then... maybe add a little more explanation. Spell things out in such a way that even an idiot like myself will understand.
Declan Davis   (talk)  12:03, 23 September 2008 (UTC)

## Line field coloring

Does anyone else think that this paper, while interesting, is not directly enough related to the subject to merit inclusion as a external link? It is already linked to from the page for Boy's Surface. A13ean (talk) 21:30, 13 July 2010 (UTC)

## Removal of section

I have removed the recently-added section on Boy's surface as redundant with the existing section Real projective plane#Immersing the real projective plane in three-space. Also, the section is mathematically incorrect: there is no embedding of the projective plane in 3-dimensional Euclidean space. Rather, Boy's surface is only an immersion. The existing section puts this all into a more appropriate context, so it seemed to me that removal of the new section was warranted, rather than attempting to correct it (and make the language, grammar, and spelling more encyclopedic). Sławomir Biały (talk) 21:18, 21 September 2010 (UTC)

## You've got to be kidding

This lead reads like what we would tell novice TA's not to say in Calculus recitations – it was a joke! The real projective plane is not that complex an object. You can use as much jargon as you like, later in the article, but a lead should give the casual reader a fighting chance. Can't someone try to do a better job than this? Bill Cherowitzo (talk) 03:12, 26 September 2011 (UTC)

I have divided up the material in the lead and shuffled it. It should read better now. Charles Matthews (talk) 08:26, 26 September 2011 (UTC)

That is certainly better, thanks, but I think more needs to be done. A description of something shouldn't really start by telling me what categories it belongs to without telling me what it is. Take that with a grain of salt, it is sometimes ok to do it when the categories are well known, but I don't think that is the case here. Bill Cherowitzo (talk) 19:28, 26 September 2011 (UTC)

## Which "diagram on the right"?

The introduction of the article ends with "...as in the diagram on the right." I am not sure which diagram this refers to. The first diagram in the article is composed of three panels. Does "on the right" refer to the rightmost panel of that diagram? I think this should be clarified. I would like to improve the wording, but first I have to know what is intended. Dratman (talk) 03:13, 13 November 2012 (UTC)

Done, but feel free to work on the wording, it could still be improved. Bill Cherowitzo (talk) 04:20, 13 November 2012 (UTC)

## Cross-capped disk

I moved across the material, which now forms the section on the Cross-capped disk, from the article on the Cross-cap. I'm not sure if there may be some errors in it, or how well it integrates with the rest of this article, so if I have messed this article up, please undo my edit and accept my humble apologies. — Cheers, Steelpillow (Talk) 11:14, 21 December 2012 (UTC)

## "Its plane of symmetry" ???

Part of the caption to the series of pictures of the unfortunately named "cross-capped disk" reads as follows:

"A cross-capped disk has a plane of symmetry which passes through its line segment of double points."

But according to the accompanying picture, this cross-cap with a disk sewn on has not one but two planes of symmetry. Yet succeeding parts of the picture captions refer to "its" plane of symmetry — implying, contrary to fact, that there is only one such plane.

For this reason, these captions need to be changed.

It would also be a very good idea to avoid trying to encapsulate the idea of a cross-cap with a disk sewn on in the shortest possible phrase ("the cross-capped disk"). As usual, the very shortest way of describing something is not the most effective way to explain what it is to someone who is not already familiar with it.Daqu (talk) 22:38, 30 January 2013 (UTC)

I agree with you, but do not have sufficient knowledge/references to know how far to change things. Still, I'm getting fed up waiting for someone more knowledgeable to fix things. I might have a go myself in a few days if I can remember to. — Cheers, Steelpillow (Talk) 20:16, 31 January 2013 (UTC)

## "glueing"

Is there a rule that my schools neglected to teach me, that monosyllables keep their e before ing if either the first letter or the letter before e is a vowel? —76.121.122.152 (talk) 18:43, 19 February 2013 (UTC)

This seems to be one of those issues where practice varies between communities (I hesitate to divide such pedantries on mere nationalistic grounds). The main rule seems to be - don't fuss about it. — Cheers, Steelpillow (Talk) 20:14, 19 February 2013 (UTC)
yeah i noticed your uk flag and churchillian userbox on your userpage. Why should us yanks fuss about being taken over? cheerio,199.33.32.40 (talk) 19:47, 13 March 2013 (UTC)

## "Topology is not concerned with flatness, and the real projective plane.."

But this is a mathematics article, and is only partly a topology article. In the manner this article is currently laid out, the remark quoted above seems out of place. Clearly a geometer could be interested in "flatness", curvature, distances, angles and other properties and other properties of various renderings of the rpp, particlularly the flat rpp embedded in R4. Just think of spherical trigonometry, etc, switched to rpp.199.33.32.40 (talk) 19:45, 13 March 2013 (UTC)

I agree that could be improved. Would it be sensible to say that "Projective geometry is not concerned with flatness..."? Otherwise, it would probably better to delete that first phrase altogether. One can understand metrics applied to the various injections (few are true embeddings) as applying to the host space and hence by association to the injected image - they are not intrinsic to the real projective plane itself. — Cheers, Steelpillow (Talk) 15:07, 15 March 2013 (UTC)
How about your phrase with the additionla words "necessarily concerned with curvature" :Projective geometry is not necessarily concerned with curvature..."199.33.32.40 (talk) 00:51, 17 March 2013 (UTC)
Done. — Cheers, Steelpillow (Talk) 13:29, 17 March 2013 (UTC)