"Every geometry is a projective geometry." So is hyperbolic geometry a projective geometry?
The great mathematician Arthur Cayley seems to have said "all geometry is projective geometry" (sorry no exact source, probably it is somewhere in Felix Klein's Erlangen program).
In projective geometry, non-intersecting lines do not exist (they all meet at infinity). In hyperbolic geometry on the other hand, through any point not on the line $\ell$, there are more than one line that does not intersect $\ell$, so at least some non-intersecting lines do exist. In all, they rather look incompatible together. A geometry cannot be both projective and hyperbolic. But still, it was Cayley who said it.
So how is hyperbolic geometry a projective geometry? I guess that it has to do with transformations/reflections and that ilk, but how does that relate to hyperbolic geometry in particular?
geometry soft-question projective-geometry hyperbolic-geometry
Ooker
WillemienWillemien
$\begingroup$ To say "all geometric is projective geometry" does not mean "every geometry is a projective geometry". $\endgroup$ – Lee Mosher Nov 5 '15 at 1:16
$\begingroup$ can you explain? $\endgroup$ – Willemien Nov 5 '15 at 9:03
$\begingroup$ What it means is that from a understanding of projective geometry one can build an understanding of a much broader swathe of geometry. For example, Euclidean geometry with all of its geometric operations (translation, multiplication or dilation, distance) can be built from projective geometry. Hyperbolic geometry, via the Klein model, can be built from projective geometry. In both of these example, models of Euclidean and hyperbolic geometry are built within projective geometry, and the axioms of Euclidean and hyperbolic geometry are proved using these models. $\endgroup$ – Lee Mosher Nov 5 '15 at 12:49
$\begingroup$ Even the real number line itself, with its operations of arithmetic, inequality, absolute value, can be built as a model within projective geometry. $\endgroup$ – Lee Mosher Nov 5 '15 at 12:53
$\begingroup$ A nice presentation of this point of view can be found in the excellent little book of @Stillwell entitled "The four pillars of geometry". $\endgroup$ – Lee Mosher Nov 5 '15 at 12:54
From the comments you included, I can see you are comparing the two geometries' synthetic axioms to see if one is a special case of the other. Of course, that is doomed to fail because the two lists contain mutually exclusive axioms about parallels (as you noticed.)
The real idea is that hyperbolic, affine and Euclidean geometries can be modeled as subsets of certain projective spaces founded upon vector spaces with suitable bilinear forms. The one for hyperbolic geometry is the Minkowski hyperboloid model, which realizes the hyperbolic plane upon the surface of a hyperboloid inside the projective plane.
Kaplansky's Linear algebra and geometry does a good job outlining all of this, although it is a little skimpy on details for a handful of topics along these lines.
rschwiebrschwieb
$\begingroup$ if they use two distinct sets of axioms, then how can hyperbolic geometry be modeled as a subset of certain projective space? $\endgroup$ – Ooker Nov 30 '17 at 4:58
$\begingroup$ @Ooker because the axioms apply to the set as a whole, and if you restrict to a subset, the "axioms of projective space" need not necessarily hold for the subset. There is nothing too mysterious about why this is possible... $\endgroup$ – rschwieb Nov 30 '17 at 11:27
$\begingroup$ ah, so we are talking about the subset of the axiom set, not of the space? So do you mean that the two share a common subset of axioms? $\endgroup$ – Ooker Nov 30 '17 at 12:35
$\begingroup$ @Ooker No, a subset of the space.. We can't throw away axioms... then we wouldn't be talking about projective space or hyperbolic space anymore. $\endgroup$ – rschwieb Nov 30 '17 at 12:44
$\begingroup$ @Ooker Did you mean that while a subset of certain projective space doesn't obey the axioms of projective geometry, it obeys axioms of hyperbolic geometry, right, the new subset of the space obeys axioms of hyperbolic geometry. You know how to extract a model of the affine plane out of the projective plane by deleting a line, right? It's similar. $\endgroup$ – rschwieb Nov 30 '17 at 14:09
The so called Klein model for hyperbolic space gives a possible answer.
Consider an ellipse in the projective plane, or equivalently a quadratic form on the 3-dimensional vector space with signature (+,+,-).
The interior of the ellipse is a model for a hyperbolic plane where the lines are the intersections of usual projective lines with the ellipse. The distance between to point (A,B) can be expressed in terms of the cross ration of (A',A,B,B') where A',B' are the intersection point of E with the projective line through A and B.
The subgroup of the group of projective transformations which preserves this ellipse (this quadratic form) is exactly the group of isometry of the interior of the ellipse endowed with this hyperbolic distance.
The same model can be generalized in any dimension, with quadric instead of ellipse and quadratic form of signature (n,1).
In a few words $PSO(n,1) \subset PSL(n+1, \bf R)$ is the subgroup of projective transformations of the projective space of dimension $n$ which preserves the interior a quadric, this interior (endowed with usual lines) is a model of the hyperbolic space.
See https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model , and reference therein).
ThomasThomas
$\begingroup$ Klein's model simply embeds hyperbolic geometry into projective geometry. $\endgroup$ – user141614 Nov 4 '15 at 9:48
$\begingroup$ @user141614: Still, though, this answers the question of what Cayley met, as opposed to the mis-quotation of Cayley in the title of this question. $\endgroup$ – Lee Mosher Nov 5 '15 at 1:15
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