Curvature of lines, curves and circle boundaries [Theory/Geometry]

[tricosahedron]

Let's take a look at straight lines (with no curvature), and curves with a fixed curvature in a hyperbolic plane:


Straight lines obviously have the lowest curvature (none).

If we increase the curvature of the line, it becomes a curve that is equidistant to a straight line, i.e. there's a constant distance between the curve and the straight line. The larger the distance between the two, the larger the curvature of the curve, and the smaller the distance, the smaller the curvature (zero distance means zero curvature, because line and "curve" overlap).

If the curvature is large enough, we get a closed circle. Larger circles have a smaller curvature, while the curvature for smaller circles is larger. As the curvature gets close to infinity, the circle becomes almost a point. The circle with the smalles curvature is a horocycle, a circle whose radius approximates infinity, but whose curvature is constant.

So, from no curvature to large curvature, we get:

Line -> equidistant curve (small distance) -> equidistant curve (large distance) -> horocycle -> large circle -> small circle -> point ("infinite curvature")


My question:
What happens if we create a curve that is equidistant to a straight line, and whose distance approximates infinity? Do we get the boundary of a horocycle, or a curve with lower curvature? In case it's the latter: How can we interpret curves whose curvature is larger than that, but smaller than the curvature of a horocycle?

Edit: I think it's the former, though:
I would assume that having two non-overlapping curves with the same constant curvature, and which are equidistant to each other, implies that their curvature has a very specific, unique value. Since this applies to two equidistant horocyclic curves, as well as to two non-overlapping curves that are equidistant to (and on the same side of) the same straight line, and whose distance to the line approximates infinity, my guess is that they're the same.


P.S.: Feel free to correct me in case I got something wrong. ;)

[Fulgur14]

Well, you made an important mistake in your very first assumption: that straight lines have no curvature.

This is not the case.

The hyperbolic space as whole has curvature (negative), and the straight lines always have the same curvature as the space they are embedded in.

This is similar to great circle on a sphere. It is the "straight line" of the sphere, but it is clearly curved.

So: straight lines have negative curvature. They actually have the "minimum" curvature -- no curve can have smaller curvature (since it's negative, "smaller" actually means "with larger absolute value" here).

Equidistant curves have also negative curvature, somewhere between the minimum curvature and zero.

Horocycles have exactly zero intrinsic curvature. We see them as curved because they are embedded in a space with even lower curvature.

Finally, circles have positive curvature, which can grow above any bounds (by having smaller and smaller circles that are more and more tightly curved). A curve with infinite curvature would have zero radius and shrink into a point.

This can be also illustrated more clearly in 3D case.

In a 3D hyperbolic space, a plane is negatively curved, and so it has hyperbolic geometry.

As you increase the curvature, the plane will become a pseudosphere (equidistant surface to a plane). Since its curvature is still negative, the geometry on the pseudosphere will be still hyperbolic. However, it will be closer to Euclidean than the geometry of plane: if you construct a triangle with the same edge length in a plane and on a pseudosphere, the inner angles of the pseudosphere one will be larger.

Let's keep one point of our pseudosphere fixed and move the plane it's equidistant to away. When we move it to infinity, the pseudosphere will become a bollosphere or horosphere, a 3D analogue of the horocycle.

The curvature of the horosphere will be zero, and that means that geometry on its surface will be Euclidean.

Horosphere has a single point at infinity. If we bring that point closer, it will morph into a sphere; the sphere has positive curvature, and its surface geometry is, of course, spherical.

[zeno]

I also think this is a horocycle. Curves of constant curvature are represented with circles or straight line segments in the Poincaré model (as used by HyperRogue). Straight lines cross the circular edge of the model at the right angle, while horocycles are tangent (equidistant curves cross at other angles, and circles do not cross the edge at all). For example, a Great Wall is a straight line or circle which crosses the edge of the model in points A and B. An equidistant going through point C is represented with a circle which goes through A, B, and C. If we go with the distance to infinity, A and B tend to the same point, so we get a horocycle.

[Fulgur14]

zeno: I didn't say it outright, but in the 3D example, it was implied. When you increase radius of circle AND when you increase the distance of equidistant curve from line, the curvature goes to zero -- so both become horocycles in the limit.

[tricosahedron]

Thanks for the correction, for an answer, and for all those helpful examples! :)

So that means the following property applies to both Euclidean and hyperbolic planes:
A line / curve that is equidistant to a straight line and infinitely distant to it, is the same as a circle boundary whose radius approximates infinity, and both are per definition a line / curve with zero curvature.

[Fulgur14]

Originally posted by tricosahedron:

Thanks for the correction, for an answer, and for all those helpful examples! :)

So that means the following property applies to both Euclidean and hyperbolic planes:
A line / curve that is equidistant to a straight line and infinitely distant to it, is the same as a circle boundary whose radius approximates infinity, and both are per definition a line / curve with zero curvature.

Pretty much. We got in a discussion with zeno about whether the lines can be said to have any intrinsic curvature -- probably not, but I think it still makes sense considering a horocycle line of zero curvature considering what happens in the 3D case.

[zeno]

Basically, the word "curvature" has many meanings. There are intrinsic curvatures which depend only on measurements inside a manifold (curve/surface/etc.), and extrinsic curvatures which say how a manifold A which is a subset of a higher dimensional manifold B is curved from the point of view of B (for example, see geodesic curvature[www.solitaryroad.com]).

What Fulgur14 said in his first post is correct when considering the intrinsic curvature (e.g. Gaussian curvature of planes, equidistant surface, horospheres and spheres in the 3D hyperbolic space). However, intrinsic curvature is defined only for higher dimensional objects, if your life is restricted to a curve and you can measure distances on it but cannot interact with the rest of the world, there is no way to tell whether that curve is a straight line, equidistant curve or horocycle.

When considering the geodesic curvature, tricosahedron was correct, for example straight lines are geodesics and thus they have geodesic curvature 0, and a horocycle has curvature 1 (assuming the hyperbolic plane has Gaussian curvature -1).