The n-Category Café

Are you sure that it’s the space of all 2d linear subspaces of $\mathbb{R}^3$ having nonempty intersection with $X$ (which is all of them) that gives you the points of Euclidean geometry?

A geometric way to say “a point $(a,b,c)$ on the cylinder determines a line $a x + b y + c = 0$” is to take the line from the origin through that point, take the plane through the origin perpendicular to it, and then intersect that plane with the plane $z=1$ (which we regard as our “Euclidean plane”). The unique line through the origin not intersecting the cylinder is $x=y=0$, which is perpendicular to the plane $z=0$, which doesn’t intersect the plane $z=1$.

The corresponding thing for a 2d linear subspace (a plane through the origin) would be to take the line through the origin perpendicular to it and intersect that line with $z=1$; then a line through the origin lies on a plane through the origin exactly when the corresponding line in $z=1$ contains the corresponding point in $z=1$. But there are plenty of planes through the origin whose orthogonal complement doesn’t intersect $z=1$ at all, namely those containing $z=0$ — yet these planes do intersect the cylinder, each in a pair of parallel lines (whereas all the other planes intersect the cylinder in an ellipse).

Of course, this is just the usual way to identify $z=1$ with the finite points of $\mathbb{R}P^2$.

I’m also confused about the hyperboloid when $s\lt 0$. Wikipedia tells me that an equation of the form

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$$

is a hyperboloid of one sheet, and that seems to be what you get here (with $a=b=1$ and $c = 1/\sqrt{|s|}$. And I can visualize the limiting process better that way too: as $s\to 0^+$ the ellipsoid stretches off to infinity in the $z$-direction becoming a cylinder, where as $s\to 0^-$ the hyperboloid gets narrower and narrower becoming a cylinder; I don’t see how to obtain a cylinder as a limiting case of a hyperboloid with two sheets.

I believe you have made a sign error. When s<0, I can try to solve 1 = $x^2 + y^2 - z^2$. I find a 1-sheeted hyperboloid. Indeed, I must, since this should be near your cylinder.

If you just want to get the space of points correct, you could instead use the quadratic form $s(x^2+y^2) + z^2$. The locus of norm 1 is then an ellipsoid degenerating to two planes at $z=\pm 1$, then becoming a 2-sheeted hyperboloid. Wrong symmetry group, though.

Your link to “Group contraction” points again to deformation theory, but I think it should point to group contraction.

David Speyer writes:

I seem to be locked out of comments at the $n$-Café. Just a quick remark:

You’ve heard of geometry on the hyperboloid with one sheet — it’s called de Sitter space! The point is that, if you restrict your signature (2,1) metric and restrict it to the tangent space to a point on the hyperboloid with one sheet, you get a (1,1) metric, so you are doing relativity, not Riemmannian geometry.

Maybe this is more complicated than you’re going for, but one context in which to make sense of all this is Cayley-Klein geometry. This may be some of the “fancy junk” that you know, but in case you don’t, and for other readers who don’t, I’ll describe it quickly.

The idea is that we fix a conic in $\mathbb{R}P^2$ and define a “geometry” on its complement (at least we get a distance measurement, an angle measurement, and a symmetry group – the projective transformations fixing the conic). For different types of conic you get different types of geometries, including hyperbolic, elliptic, Euclidean, and Lorentzian.

The sneaky bit is that you have to be very general about what you mean by a “conic”. All ordinary conics in $\mathbb{R}P^2$ are equivalent, having a homogeneous equation like $x^2+y^2-z^2=0$. When points of $\mathbb{R}P^2$ are regarded as 1d subspaces of $\mathbb{R}^3$, then this conic is a double cone through the origin, and its complement is disconnected into the “inside”, which is the lines intersecting the two-sheeted hyperboloid (in a pair of antipodal points), and the “outside”, which is the lines intersecting the one-sheeted hyperboloid (also in a pair of antipodal points). The inside geometry is hyperbolic, while the outside geometry is hyperbolic along some lines (those that intersect the conic) and elliptic along others (those that don’t).

In addition to ordinary conics, however, we also allow complex conics without any real points, typified by the homogeneous equation $x^2+y^2+z^2=0$, which has no solutions in $\mathbb{R}P^2$ (since $(0,0,0)$ is not a homogeneous coordinate) but has plenty in $\mathbb{C}P^2$. We still however consider the geometry as defined on the complement of the conic in $\mathbb{R}P^2$, not $\mathbb{C}P^2$, so in this case we have all the points of $\mathbb{R}P^2$, and we get elliptic geometry.

We also allow degenerate conics; but we have to be extra sneaky about what we mean by those. A nondegenerate conic determines, and is determined by, its “dual conic” consisting of all the lines tangent to it. However, for degenerate conics this relationship breaks down, and it turns out that for the purposes of Cayley-Klein geometry we should consider “a conic” to be a pair consisting of a “point-conic” and a “line-conic” that are “related like duals”. In the nondegenerate case, each determines the other, but not in the degenerate case.

A degenerate point-conic can be a pair of lines ($x^2-y^2=0$) or a single “doubled” line ($x^2=0$). Dually, a degenerate line-conic can be a pair of points or a single “doubled” point. And all of these could be either real or imaginary. It turns out that the pair consisting of a single real “doubled” line and two imaginary points on that line gives rise to Euclidean geometry. The space of points is the complement of one line in $\mathbb{R}P^2$, where the missing line is of course “at infinity”. Unlike in the hyperbolic and elliptic cases, in this case (sometimes called “parabolic”) we can’t define an absolute distance, but with a limiting process we can compare relative distances. Accordingly, the transformation group preserving the line at infinity and the two fixed imaginary points on it is the group of similarity transformations of Euclidean geometry.

I don’t have time to think about it myself right now, but is section 3.2 of this paper possibly relevant?

A quick point: The group of matrices preserving a singly-degenerate quadratic form on $\mathbb{R}^3$ is four-dimensional, containing not only the $xz$-shears, the $yz$-shears, and the $xy$-rotations, but also the $z$-scaling.

The group preserving a doubly-degenerate quadratic form is six-dimensional, containing the expected $zx$- and $zy$-shears, as well as the $xy$- and $yx$-shears, the $x$-scaling, and the $y$-scaling.

These symmetries can be mapped to the other model, where they all affect the metric. However, scaling in the $z$-direction is projectively equivalent to isotropic scaling in the $xy$-plane, so that particular combination does not affect the metric.