Talk:Real projective line

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New start[edit]

Here is my contribution for a dedicated article: "Real projective line".Rgdboer (talk) 20:55, 21 June 2015 (UTC)

Thanks for great work! Tkuvho (talk) 07:32, 22 June 2015 (UTC)


The current version contains the footnote "If a real projective line happens to appear in a non-Desarguesian plane the harmonic structure cannot be presumed." Now the usual embedding of this would be in the real projective plane, rather than some other non-Desarguean plane. This is certainly possible for example by using the projective plane over the Cayley numbers. However, I am wondering about the relevance of this footnote at this page. Tkuvho (talk) 07:32, 22 June 2015 (UTC)

The embedding in a non-Desarguesian plane is but one possibility. If the usual axioms are applied to the line with no embedding, the group of motions can be considerably larger. So if the article is to mention the non-Desarguesian direction, this requires a more thorough covering the possibilities. —Quondum 13:24, 22 June 2015 (UTC)
The embedding of a real projective line in a non-Desarguean plane has no projective meaning, strictly speaking. Of course in differential geometry it does, but it is not the subject of this page. Tkuvho (talk) 14:37, 22 June 2015 (UTC)
I agree, that kind of embedding is not natural and mentioning non-Desarguesian planes seems quite out of place in this article. Bill Cherowitzo (talk) 17:39, 22 June 2015 (UTC)

Construction over the real line: can't be glossed over[edit]

The construction over the real line, recently added, needs considerable work. One cannot simply add a point to the real line as defined in that article and get the real projective line as a geometric object. That article gives a hopelessly unclearly defined object for the purpose. One needs to start specifically with the real affine line, a homogeneous space, whereas it is difficult, from that article, to think, geometrically, of anything other than an object of which the group of automorphisms is the trivial group. Secondly, even starting with a one-dimensional affine real space (also not Euclidean: it must have no metric structure), one has to include a construction for changing the added point into a normal point, i.e. expanding the group of motions in a particular way. So, placing this as the first alternative of a definition for the construction implies far too much assumption; this needs to be fixed. For now, I'd suggest removing this until a full section explaining the construction correctly can be added. —Quondum 13:44, 22 June 2015 (UTC)

The only problem is that all of the theorems of projective geometry such as Pappus, Desargues, Pascal, etc. become inapplicable to affine configurations if one does not include a mention of the construction via adding points at infinity, which is a considerable loss if one is interested in actual applications of projective geometry :-) Metric structures are irrelevant here. Tkuvho (talk) 14:39, 22 June 2015 (UTC)
Perhaps thinking of the construction of the real projective line as the set of all lines through the origin in R2 would be fruitful. Identifying these lines with their slopes gives the real line and the vertical line through the origin gives the point at infinity. The homogeneous nature of the result is quite apparent from this viewpoint. The non-uniqueness of the coordinatization (or embedding, if you prefer) can also be made very clear. Bill Cherowitzo (talk) 17:34, 22 June 2015 (UTC)
This view of the construction also shows that the first sentence in the definition section is not really correct. I would also advocate putting in a mention of the point at infinity, but this has to be done carefully. One should not start with the real line and think of adding a point to it, as this gets you into the difficulties that Quondum has mentioned (not insurmountable, but it requires some work to get it right). Rather, start with the real projective line and point out that the removal of one point leaves you with a copy of the real line and the removed point can be thought of as the "point at infinity" with respect to that copy. Bill Cherowitzo (talk) 16:56, 23 June 2015 (UTC)

Day-one review[edit]

Good participation for mid-summer. All comments directed to improved article with clear experience in bringing this topic out for general review. Some changes were made this afternoon reflecting discussion in the last 24 hours, including a link to Point at infinity showing that it is relative to chart selection. Please add to the See also as appropriate; my contribution is slope. This subtle little object from an old geometrical practice has an important place in math, we do well to explicate it clearly. As for the non-Desarguesian situation, that technicality that Hilbert and others used to upset expectations of old, it is mentioned as a caveat since it may arise in advanced studies.Rgdboer (talk) 23:25, 22 June 2015 (UTC)

Mathematics made difficult[edit]

This is probably the most complicated way I have ever seen to describe a circle. The lede says that the thing is homeomorphic to a circle, seemingly implying that it has a different geometric structure, and then gratuitously mentions that it is a non-trivial smooth manifold. Actually, though, the natural distance function on any real projective space is simply the angle between lines, which can be up to \frac{\pi}{2}. In this case, this is the same as the metric of a circle. So it's not just homeomorphic to a circle, it is a circle, specifically a circle of radius \frac{1}{2}. Will anybody mind if I simplify the lede accordingly? Also the construction itself is a little technical and is not intuitively explained - it should be stated explicitly that the points of \mathbb{R}P^{1} are lines in \mathbb{R}^2 passing through the origin. And we should definitely mention that it is commonly understood as the one-point compactification of the real line; even though, as Quondum notes, this does make it a little less clear what the metric is, we can do away with this problem by mentioning that the metric is that of the circle. (A little more technically, the Riemannian metric of the real-line-with-point-at-infinity model of \mathbb{R}P^{1} is \frac{dx}{1+x^2}, which is obtained by pushing forward the Euclidean metric dx by \arctan, the obvious map which takes points on the real line to angles.)

In the "automorphisms" section, it doesn't quite make sense to say "the mappings are homographies" - what we really mean is that the homographies are mappings of special interest. We also should probably mention that PGL_{2}(\mathbb{R}) is the isometry group of the hyperbolic plane, which in the Poincare disc model can be seen as the interior of the circle. The real projective line then can be seen as imbedded in the complex projective line, and this explains the statement that the group of homographies is intermediate between the modular group and the full Moebius group. Does anybody mind if I make these changes as well?

I think the "Structure" section might do with some expansion - I really don't understand what it says and I'm not familiar with all that classical geometry stuff. --Sammy1339 (talk) 05:58, 23 June 2015 (UTC)

This comment expresses some legitimate concerns. There is one point I don't follow. It is true that the homographies are isometries of the hyperbolic plane, but why do you write "the real projective line then can be seen as imbedded in the complex projective line"? The imbedding in the complex projective line is merely the extension of coefficients, and needs not be explained in terms of hyperbolic isometries. Tkuvho (talk) 07:35, 23 June 2015 (UTC)
I guess that the homographies of the real projective line may be related to the motions of the omega points of the hyperbolic plane; this would explain the group being the same and may be worth a mention. This seems far more natural (or at least more intuitive) to me than the relationship with the complex projective line; Tkuvho's argument around on complexification makes sense. —Quondum 14:53, 23 June 2015 (UTC)
Well the idea is that you take \mathbb{C}^2 and form the complex projective line, and look at what happened to the plane spanned by the two real axes: it became a real projective line which is the boundary of the upper half-plane H^{2}. The homographies of this line are precisely the Moebius transformations which fix it - these are the linear fractional transformations with real coefficients; they are also the isometries of H^{2}. This explains the relationship mentioned in the text, that the modular group (a group of hyperbolic isometries) is a subgroup of the group of homographies of \mathbb{R}P^1, which is in turn a subgroup of the Moebius group. --Sammy1339 (talk) 16:20, 23 June 2015 (UTC)
There is no connection whatsoever in what you describe between the representation of the real projective line as a subset of the complex projective line, on the one hand, and its representation as the "boundary" of the hyperbolic plane, on the other. Tkuvho (talk) 16:24, 23 June 2015 (UTC)
Of course there is - the hyperbolic plane (half-plane model) is half of the Riemann sphere a.k.a. \mathbb{C}P^1, and the \mathbb{R}P^1 constructed as above is the boundary of it. It's not a coincidence that the Moebius transformations (complex homographies) which fix H^{2} (or equivalently which fix \partial H^{2}) are precisely the hyperbolic isometries. This occurs because the Moebius transformations are the conformal bijections which map circles to circles, and the geodesics of H^{2} are circles which intersect the boundary orthogonally, so a transformation will be an isometry of H^{2} iff it fixes \partial H^{2}, maps circles to circles, and preserves angles. --Sammy1339 (talk) 16:56, 23 June 2015 (UTC)
This is a misleading point of view. The complex homographies in general don't fix the real projective line, and form a much larger group, namely PSL(2,C), than the real homographies. Not much is gained by viewing the hyperbolic plane as "half" the complex projective line. Tkuvho (talk) 17:06, 23 June 2015 (UTC)
Right, so the point is that those elements of PSL_{2} (\mathbb{C}) which do fix the real line are the real homographies. This is the way to find out that the isometry group of the hyperbolic plane is PSL_{2} (\mathbb{R}) - I don't know of another way to prove this, offhand. It's also the view universally taken in the study of Riemann surfaces and analytic number theory. --Sammy1339 (talk) 17:19, 23 June 2015 (UTC)
But the hyperbolic metric has little to do with this, and it certainly cannot be induced by the imbedding into the complex projective line, so in a way using the phrase "hyperbolic plane" in relation to the complex projective line is neither here nor there. Fractional linear transformations act in the upperhalf plane and give all the isometries, so one doesn't need to ever speak of projective geometry to understand basic facts about the hyperbolic group of isometries. Of course the crossratio provides a nice way of defining the hyperbolic metric. Tkuvho (talk) 17:26, 23 June 2015 (UTC)
I see what you mean. But everyone says this, and it's natural for two reasons. One, because the hyperbolic metric is the one given by the uniformization theorem. Two, you can look at it the other way around: try to find a metric on the upper half plane for which PSL_{2} (\mathbb{R}) is the isometry group, and you will discover the hyperbolic metric. It's also useful to remember where "linear fractional transformations" came from - they were once linear endomorphisms of \mathbb{C}^2 which acted as (z,w)\mapsto (az+bw,cz+dw) and then after projectivization we sent (z,w) to (z/w,1) and (az+bw,cz+dw) to (\frac{a(z/w)+b}{c(z/w)+d}, 1). So saying we "don't need" the \mathbb{C}P^{1} picture because we can just define linear fractional transformations ad hoc and discover by inspection the miraculous coincidence that they form the group PSL_{2}(\mathbb{C}) is, in my view, sort of a cheat. --Sammy1339 (talk) 17:51, 23 June 2015 (UTC)


Today the differential geometers' definitions were quoted to show their uniformity and presumption of a Euclidean sphere to make the definition. Meanwhile an editor has proposed a definition presuming knowledge of vector spaces and deriving equivalence from that structure. Surely mathematically minded people look at foundations of their concepts and we should do so here. The differential geometers are pragmatic but unconcerned about the projective structure that might result from fields besides R. Similarly, appealing to vector space knowledge may be distracting. That is why the original definition given two days ago (a purely algebraic construction) is to be preferred.Rgdboer (talk) 23:21, 23 June 2015 (UTC)

The previous definition has been kept. Given alone, it is misleading, as it suggests wrongly that some points of the projective line are special, such as the two points that have a zero in their projective coordinates. Moreover, the concept of vector space is purely algebraic. On the other hand, I cannot imagine that anybody may learn more than trivialities in differential geometry, without knowing of vector spaces. In fact a tangent space is a vector space and almost nothing can be done without this structure.
Nevertheless something was lacking in the previous definition (even with my edit); it is the fact that the definition of the projective line as a quotient space must include the definition of the topology and of the differential structure as those of a quotient spaces. I'll try to clarify that. D.Lazard (talk) 07:50, 24 June 2015 (UTC)
My reaction as well to the comment is that the algebraic statement is still the opening one, albeit in a more general form, which IMO is a distinct improvement. —Quondum 12:38, 24 June 2015 (UTC)

Is the projective line a circle?[edit]

The new lead asserts that the real projective line is a circle. This wrong. The real projective circle is homeomorphic to a circle, but it is not, as far as I know diffeomorphic to a circle. One may see that by observing that the universal cover of the projective line is not the projective line itself, but a projection of index two of a circle. Also, there is no embedding of the projective line into a Euclidean plane.

Moreover, the new lead does not have a neutral point of view, by privileging the topological point of view with respect to the other points of view (geometric, algebraic, differential, analytic, ...). For these reasons, I'll revert this change of the lead.

On the other hand, I am not satisfied by the previous state of the lead, and I will edit it, but only when the body of the article will be sufficiently expanded, for having a lead reflecting the body of the article. In fact the following sections are yet lacking: Projective frames and change of homogeneous coordinates, homographies, cross-ratio, ... D.Lazard (talk) 16:12, 24 June 2015 (UTC)

No, the real projective line is just a circle. There are no competing points of view here - \mathbb{R}P^{1} with the usual metric is isometric to S^{1} with its usual metric, and there is no difference between homeomorphism and diffeomorphism in dimension less than four. See my comments in the section "Mathematics made difficult" above. --Sammy1339 (talk) 17:04, 24 June 2015 (UTC)
The map that sends a point of the circle to the line passing trough the point and the center of the circle defines a map from the circle onto the projective line. This map is locally a difffeomorphism (it is even analytic), but globally, it is not a diffeomorphism, as this map is not one to one. If you can provide explicitly a global diffeomorphism between the circle and the projective line, I will reconsider my position. I doubt that such a diffeomorphism exists, because the property of being identical to its universal cover is invariant by diffeomorphism, and the circle has this property and the projective line has not (its universal cover is just defined by above map. D.Lazard (talk) 18:21, 24 June 2015 (UTC)
I think the source of confusion here is that, for n \geq 2, the universal cover of \mathbb{R}P^{n} is S^{n}. Clearly this must fail for n=1, because the fundamental group of any universal covering space must be trivial, and \pi_{1}(S^{1})=\mathbb{Z}. The explicit diffeomorphism from \mathbb{R}P^{1} to S^{1} takes the line of elevation \theta above the real axis and sends it to the point on circle of radius \frac{1}{2} centered on 0 located at angle 2 \theta above the real axis. You can check it's an isometric bijection of Riemannian manifolds. --Sammy1339 (talk) 18:43, 24 June 2015 (UTC)
A point to consider is that any projective line has structure in the form of a group of motions. Topological objects such as circles do not have the given automorphism group. And there is no such thing as "the usual metric" on the real projective line. —Quondum 19:59, 24 June 2015 (UTC)
Okay, by the "usual metric", I meant that of angles between lines. To be super explicit, a global diffeomorphism from \mathbb{R}P^{1} to S^{1} is defined by
\{ (x,y) \in \mathbb{R}^2 |y=\lambda x, \lambda \in \mathbb{R} \} \mapsto \frac{1}{2} e^{2i\arctan{\lambda}}
\{ (x,y) \in \mathbb{R}^2 |x=0 \} \mapsto -\frac{1}{2}
So these two spaces are identical in every sense. \mathbb{R}P^1 does not "have" the automorphism group PSL_{2}(\mathbb{R}), in any way that S^{1} doesn't. They are the same manifold. --Sammy1339 (talk) 20:27, 24 June 2015 (UTC)
You're losing me. Not that it is important. The entire group of motions of RP1 is PSL2(R). But the auto-diffeomorphisms on S1 is a larger group. They are topologically equivalent, but not equivalent as geometries. —Quondum 00:04, 25 June 2015 (UTC)
So, someone has made up the idea that the real projective line "has the automorphism group" PSL_{2}(\mathbb{R}). We say that S^{2} has the automorphism group PSL_{2}(\mathbb{C}) because the action of this group preserves the conformal structure. But in \mathbb{R}P^{1} there are no angles. In higher dimensional real projective spaces, homographies map lines to lines. But in \mathbb{R}P^{1} there are no lines. What structure of \mathbb{R}P^{1} is preserved by the action of PSL_{2}(\mathbb{R})? Of course you could make one up. But even if you do, you can put the same structure on S^{1} because they are two words for the same thing.
Actually the only thing interesting about the action of PSL_{2}(\mathbb{R}) on S^{1}, as far as I'm aware, is that, if S^{1} is viewed as the boundary of a hyperbolic plane, it induces hyperbolic isometries. I don't know of any intrinsic structure on S^{1} that this invariant under this action. --Sammy1339 (talk) 00:31, 25 June 2015 (UTC)
What you're saying is greek to me. PSL2(R) is the group of motions of the real projective line whose points are defined as equivalence classes over a two-dimensional real vector space. This is hardly uninteresting, and has nothing to do with hyperbolic geometry. But since we are clearly missing each other, I don't see the point of continuing this. —Quondum 00:46, 25 June 2015 (UTC)
Well, what do you mean by "the group of motions?" That term implies there is some structure invariant under these motions. But there isn't. --Sammy1339 (talk) 00:49, 25 June 2015 (UTC)
The cross-ratio. —Quondum 01:40, 25 June 2015 (UTC)

Sorry I was wrong about everything I was saying about differentiable geometry: The projective line and the circle are diffeomorphic (that is isomorphic as differentiable manifolds) and even birationally equivalent as algebraic varieties (I believe, but I am not sure, that they are not isomorphic as real algebraic varieties). But this is not the question. Both standard definitions of the circle and the projective line make them naturally homogeneous spaces for some group, but not the same group. For the circle, this is the group of the plane rotations. For the projective line, this is the group of homographies PSL2(R), whose action is induced by the action of GL2(R) on the lines. Note that when the projective line is defined through synthetic geometry, one gets exactly the same homography group.

Thus, if the circle and the projective line are isomorphic as manifolds, they are very different mathematical objects. Beginning the lead by "the real projective line is a circle" is as ridiculous (and confusing for the layman) as beginning Parabola article by "the parabola is a line" (Sammy1339's arguments apply exactly in the same way to this case). D.Lazard (talk) 09:16, 25 June 2015 (UTC)

There are two relevant differences between a parabola and a line:
1. The parabola is a specific embedding of a line into Euclidean space; it is not defined intrinsically.
2. The natural metric on a parabola is the one induced by restricting the Euclidean metric. With this metric, the parabola is homeomorphic but not isometric to the Euclidean line. It has curvature.
It has been said that the real projective line is a homogeneous space for the group of homographies; this is not true. In general, P(V) is a homogeneous space for the projective orthogonal group PO(V).
As Quondum points out, the action of PSL_{2}(\mathbb{R}) preserves the cross-ratio. In order to define the cross-ratio, however, we need to use the metric. This metric makes \mathbb{R}P^{1} a circle of radius \frac{1}{2}. Isometric manifolds have all the same properties; they are the same manifold. To say that this group action preserves the cross-ratio on circles does not imply any difference between \mathbb{R}P^{1} and a circle; it's just a true fact about circles.
\mathbb{R}P^{1} does not have any structure that a circle of radius \frac{1}{2} does not have. This is because it is (isometric to) a circle of radius \frac{1}{2}. --Sammy1339 (talk) 15:38, 25 June 2015 (UTC)
The cross-ratio is a purely algebraic concept that is defined over any field, even finite ones. There is no need of any metric to define it.
Isometric manifolds have all the same properties; they are the same manifold. I deduce that two different circles of the same radius in Euclidean plane are the same circle :-)
I could answer more, but this seems unnecessary. D.Lazard (talk) 15:56, 25 June 2015 (UTC)
Of course I mean intrinsically defined isometric manifolds are the same, as I thought I made clear. I'm not getting the bit about the cross-ratio not requiring a distance function. If you have four points on a circle a topological S^{1}, how do you define their cross-ratio without referring to their distance from eachother? --Sammy1339 (talk) 16:04, 25 June 2015 (UTC)
As far as I know, cross-ratio is defined only for lines and conic sections. Do you have a reference for a definition for other curves? If not, Wikipedia is not the place for discussing this issue. D.Lazard (talk) 21:59, 25 June 2015 (UTC)
Okay... so you are saying that if four points on this object have a cross ratio, it must be a conic section, presumably a circle? Then what is the remaining objection to calling it a circle? --Sammy1339 (talk) 00:32, 26 June 2015 (UTC)
It is your right to call an hyperbola "circle", "apple" or "mountain", but it is not the problem of Wikipedia. D.Lazard (talk) 07:33, 26 June 2015 (UTC)


(Copied from preceding thread) Both standard definitions of the circle and the projective line make them naturally homogeneous spaces for some group, but not the same group. For the circle, this is the group of the plane rotations. For the projective line, this is the group of homographies PSL2(R), whose action is induced by the action of GL2(R) on the lines. Note that when the projective line is defined through synthetic geometry, one gets exactly the same homography group. D.Lazard (talk) 09:16, 25 June 2015 (UTC)

You note that the same homography group arises from the synthetic axioms. Since the axiom of at least two distinct lines must be dropped for the line to exist, the homography group would seem to be the group of collineations, which is Sym(R ∪ ∞). The alternative of defining the line in terms of an embedding in a higher-dimensional space seems unnatural, since the same would apply to the plane. —Quondum 13:38, 25 June 2015 (UTC)
For projective lines, the collineation group is not Sym(R ∪ ∞), as explained in Collineation § Collineations of the projective line. Here, we have the field of the reals that has no non-trivial automorphism. It follows that, here, homographies and collineations are exactly the same thing. In real synthetic geometry, homographies of the line are usually defined as restriction to the line of collineations or homographies that fix globally the line, in a projective space of higher dimension. What I was saying, is that this definition of homographies is exactly the same as the one deduced from linear automorphisms of the vector plane of dimension 2. IMHO, the only questionable assertion of the new lead, possibly requiring a source, is the implicit assertion that elements of Sym(R ∪ ∞) are qualified of geometric transformations only if they are homographies. I am pretty sure that this is also implicit in almost every text talking of geometric transformations. D.Lazard (talk) 15:45, 25 June 2015 (UTC)
You seem to have jumped away from the synthetic context that I was referring to. In synthetic geometry, one works with the primitives of points, lines and their incidences. Your link refers to vector spaces and semilinear products. In the section just above, we see: "For dimension one, the set of points lying on a single projective line defines a projective space, and the resulting notion of collineation is just any bijection of the set." The group of set bijections of a set SS is Sym(S). —Quondum 23:17, 25 June 2015 (UTC)
I have not "jumped": It has been proven in the beautiful book Geometric algebra by Emil Artin that projective geometry defined from linear algebra, and synthetic projective geometry with Pappus's and Desargues' theorem are strictly equivalent (I do not know if this book is really the first proof of the equivalence of the two approaches). Thus every assertion stated or proved in one context is valid in the other context. On the other hand, it is clear that in the case of lines, two different definitions of "collineation" coexist, the etymological one (preserving alignment), and the "restricted" one (restriction of a collineation in higher dimension, or, equivalently, composition of a field automorphism and a homography). In the case of this article, as the real field has no non-trivial automorphism, there is no need to talk about "collineations". I do not see the problem of keeping (in synthetic context as well as in linear algebra context) the meaning of "homography", which has been introduced a long time before axiomatization of synthetic geometry and the introduction of linear algebra in this field. D.Lazard (talk) 08:28, 26 June 2015 (UTC)
Part of the problem here is that we have not clarified which axioms apply. The fundamental synthetic axioms of geometry (excluding taking the "theorems" of Desargues and Pappas as axioms) lead to non-desarguesian planes, and these are an area of study in synthetic geometry. Similarly, Desarguesian non-Pappan geometries are also studied and are well-behaved and fit with the treatment via homogeneous coordinates. Thus, to assume that in the context of synthetic geometry these axioms automatically apply is not valid. Furthermore, if the line is not axiomatically embedded in a higher-dimensional space (e.g. defined as a line within a two-dimensional geometry), but instead an axiom is dropped to allow a line to exist as an independent geometric object, then I expect that no axioms based on incidence will induce the standard homographies on any set. I have already made reference to the embedding requirement being significant. —Quondum 13:55, 26 June 2015 (UTC)

Apparently, this discussion did not start from the article content, but from a side remark by myself in this talk page. For the moment, "homography" appears only as a single word in the lead, and in the last section. Synthetic geometry is not cited explicitly, but only implicitly, in the last section, through the words "central projections" and "parallel projections". This last paragraph, which I have never edited, says essentially the same as me, namely that classical geometry define the same homographies on the line as the modern presentation through linear algebra. IMHO this does not derserve to be discussed further, before having in the article a detailed definition/description of the homographies of the projective line. Then we could discuss how to describe the relationship between the classical definition and the modern one through linear algebra. D.Lazard (talk) 14:44, 26 June 2015 (UTC)

Yes, that's a fair comment. —Quondum 15:59, 26 June 2015 (UTC)

Several comments. The circle and RP1 are diffeomorphic. They are also isomorphic as algebraic varieties. That does not mean that they are "the same". The circle is a particular conic section, for instance. It is a special case of an ellipse, for instance, any while every circle is isomorphic to every ellipse as an algebrair variety, it is not true that every ellipse is a circle. The real projective line, as the set of one dimensional linear subspaces of a two dimensional vector space over the reals, does not support a "natural" metric. It is the homogeneous space PGL (2)/B, where B is a borel subgroup. This is the Poisson boundary of the associated RiemannIan symmetric space PGL (2)/K, where K=SO (2) is the maximal compact subgroup (this is the hyperbolic plane). The real projective line naturally carries a projective structure, that is, a maximal atlas whose transition functiond are fractional linear. This can equivalwntly be given as a Cartan connection on a PGL (2) bundle over RP1 I. In contrast, the circle does carry an invariant Riemannian metric. Indeed, it is the unique one dimensional Riemannian symmetric space of compact type, which is a principle homogeneous space for the group SO(2). Sławomir Biały (talk) 23:01, 26 June 2015 (UTC)

embedding link target?[edit]

The link embedding of a real line in a projective line (on this change) seems incorrect to me: an embedding is an isomorphism, which requires all the structure of the embedded object to be present and to be preserved in the space in which it is embedded, unless the isomorphism is specified to be on a restricted structure. In particular, the the real line has a ring structure, whereas the image of the embedding does not (though it may be considered to unnaturally induce one). And even a geometric affine line has a problem in being embedded in a projective space in this sense: the groups of motions is different. Is there better a way of expressing this? It is really just an injective map, though there is clearly some structure that is preserved, such as the natural topology (which was why I selected that particular link, though it might not be a good one). —Quondum 03:23, 27 June 2015 (UTC)

It's an embedding of the affine real line. The stabilizer of the point at infinity is the affine group of the line. Sławomir Biały (talk) 04:07, 27 June 2015 (UTC)
I can almost see that. The affine real line with group of symmetries being the stabilizer of one point on the projective real line has a smaller group off symmetries than the line it is to be embedded in, and this is equivalent to more structure than embedding space has. So technically my point still stands: the embedding is not a true embedding of the whole object, but of an object with some of its structure "forgotten", so it makes no difference whether we start from the field of real numbers or the affine real line. But this point is rather subtle, and unless there is some term such as a "forgetful embedding", it probably cannot be clarified in the article by choice of terminology. So I don't see that there is much to be changed. —Quondum 14:51, 27 June 2015 (UTC)
There's always "more structure", for any embedding, because then one doesn't just have a space, but a space with another embedded inside it. In this case, the subgroup that preserves the embedded submanifold is the affine group of the line. The symmetry group of the embedded affine line is as large as it can be, as a subgroup of PGL(2). Sławomir Biały (talk) 17:39, 27 June 2015 (UTC)
Yes, I was thinking of that exact example before your response. Anyway, so it comes down to that it is a true embedding and my challenge does not hold, and it is the definition of embedding that is not clear to me. The article Embedding might do with a little elaboration. I probably had the induced structure argument back-to-front. —Quondum 20:30, 27 June 2015 (UTC)