# parallel postulate of Euclidean geometry and curvature

In elementary geometry, we have two standard examples which violate the (strong) parallel postulate of Euclidean geometry: in hyperbolic geometry, we have more than one parallel through a point which doesn't lay on a given line, and in spherical geometry, we have no parallels at all.

In both of these geometries, we have some kind of curvature: if we draw a triangle in Poincaré's disc model of hyperbolic geometry, the sides of a triangle can be bent inward and in spherical geometry, the sides of a triangle are bent outward.

My question is: does the absence of the (strong) parallel postulate imply a geometry with curvature and does the presence of the parallel postulate imply the absence of curvature? I know that this question may sound really vague. It is motivated by playing with Euclids axioms and I don't have a precise mathematical definition of curvature in mind. Also, I don't have knowledge of differential geometry. I hope it is still possible to answer it somehow.

## 2 Answers

This doesn't directly answer your question, but should help bridge the gap between Euclidean geometry and curvature, and so may be of interest nonetheless.

In the sense you're asking, "curvature" (meaning Gaussian or intrinsic curvature) can be defined using angular defect in small triangles. More precisely, if $\triangle ABC$ has "straight lines" (geodesic segments) as sides, the sides of $\triangle ABC$ enclose a topological disk, and if the interior angles are $\alpha$, $\beta$, and $\gamma$, then the total curvature inside $\triangle ABC$ is the angular defect $(\alpha + \beta + \gamma) - \pi$. (This fact is a special case of the "local Gauss-Bonnet theorem".)

Particularly, a surface has curvature identically zero if and only if every sufficiently small geodesic triangle has total interior angle $\pi$.

To pass from local geometry (interior angles of triangles) to global assertions such as the parallel postulate, we need global hypotheses. Commonly, one assumes the surface is "geodesically complete" (geodesics can be extended indefinitely in both directions), has constant curvature (the angular defect of a geodesic triangle is proportional to the triangle's area, and the curvature is the constant or proportionality), and is "simply-connected" (every closed loop in the surface can be shrunk continuously to a point within the surface). It turns out precisely three surfaces satisfy all three conditions: The Euclidean plane, the sphere, and the hyperbolic plane.

If we assume only completeness and constant curvature, infinitely many possibilities open up:

• Circular cylinders and flat tori (see below);

• The real projective plane;

• Hyperbolic surfaces.

In hyperbolic surfaces other than the hyperbolic plane, there exist self-crossing geodesics, so the very notion of a line is complicated. In the other surfaces, every geodesic is either closed (returns to its starting location after finite distance with its initial direction, like a great circle on a sphere or a latitude line on a cylinder) or does not self-intersect.

A complete surface $S$ whose geometry satisfies the strong parallel postulate (and none of whose geodesics are closed) is the Euclidean plane: Pick a point $O$ arbitrarily, and pick lines $X$ and $Y$ through $O$ that meet perpendicularly. By the parallel postulate, for each point $x$ on $X$, there is a unique line through $x$ and parallel to $Y$, and similarly for each $y$ on $Y$ there is a unique line through $y$ parallel to $X$. This "Cartesian graticule" sets up an isometry between $S$ and the Cartesian plane with the Euclidean metric.

There are flat (curvature-zero) surfaces that satisfy the strong parallel postulate but are not the plane. A circular cylinder in space is the easiest to visualize. A "line" on a cylinder is a generator (a line parallel to the axis), a helix, or a latitude (a circle lying in a plane orthogonal to the axis).

There are also "compact" examples, namely tori, which may be viewed as the universe of a two-dimensional video game in which exiting the screen causes one's avatar instantly to reappear on the opposite edge of the screen. A "line" on a flat torus is either a closed loop that "winds" some integer number of times left-to-right and some (relatively prime) integer number of times bottom-to-top, or else is an irrational winding passing arbitrarily close to every point of the torus.

Since we're delving into fine points, a torus is topologically a product of circles, but not every flat torus is a "Riemannian product" or "rectangular". That is, in a flat torus there always exists a shortest closed geodesic $\gamma_{0}$, but a geodesic meeting $\gamma_{0}$ perpendicularly may not be a closed curve at all (much less a "short" closed curve).

First of all, you have to decide which axioms of euclidean geometry you want to keep in the Riemannian setting. The most natural thing to do is to work with simply connected complete noncompact Riemannian surfaces. (Cf. Andrew Hwang's answer.) The Riemannian notion of a line is a complete geodesic. You probably also want to assume that two distinct lines intersect in at most one point. The most natural way to achieve this is to assume that the Riemannian metric has no conjugate points. Now you have to decide if your Riemannian surface $X$ has any symmetries (isometries) and if it does, how many. In the (non)euclidean geometry the group of symmetries is assumed to be quite large: It acts transitively on the unit tangent bundle. Assuming this is too much since this leaves you only with surfaces of constant curvature. You can settle for existence of a discrete group of symmetries $G$ such that $X/G$ is a compact surface $S$. Once you assume this, the situation again becomes quite rigid, but in a less obvious fashion:

1. If $S$ is a torus of a Klein bottle, then by a theorem of Hopf (generalized by Burago and Ivanov in all dimensions) the Riemannian metric is flat and you conclude that $X$ is isometric to the euclidean plane.

2. If $S$ has negative euler characteristic (say, oriented surface of genus $\ge 2$) then with a bit of work you conclude that $X$ violates the 5th postulate.

Now, if you do not make any assumptions about the group of symmetries, then pretty much nothing is known. For instance, the following is an open problem which I learned from Keith Burns:

Problem. Suppose that $X$ has nonpositive curvature and satisfies the 5th postulate. Is it true that $X$ has zero curvature, i.e. is isometric to the euclidean plane?