Invariance of ball size under translation on the pseudosphere

Question

On the pseudosphere, do the volume of metric balls and area of metric spheres change with the centerpoint? (I'm asking because they don't on the sphere.)

Answer 1

I don't think it makes sense to think of the tractricoid as "the complete surface of constant negative curvature", for two reasons.

One: The tractricoid has the singular points at the equator, which means that it is not an honest smooth Riemannian surface per se. In particular, the tractricoid of sectional curvature $-1$ is not isometric to any quotient the hyperbolic plane (which is a complete surface of constant sectional curvature $-1$).

Two: Even among (smooth) Riemannian manifolds, there is no unique complete surface of constant sectional curvature $-1$. (The hyperbolic plane is the unique simply connected one, but it has many interesting nontrivial quotients.)

The sphere has the nice property that for any two points $p$ and $q$, there is an isometry of the entire sphere taking $p$ to $q$. (It's a "homogeneous space".) Your result about the areas of metric discs follows instantly. The hyperbolic plane also has this property. The tractricoid does not (its isometries are only the rotations about the axis and the flip across the equator). For this reason (among many others), the hyperbolic plane is the "correct" generalization of the sphere to negative curvature.

As Wikipedia points out, the tractricoid is locally isometric to the hyperbolic plane (away from the singular locus). More generally, any two points on any two surfaces of equal constant negative sectional curvature have isometric neighborhoods. This is a standard result in Riemannian Geometry, not completely trivial to prove, which applies to higher-dimensional manifolds as well; for instance, it follows from Corollary 11.13 in Lee's book Riemannian Manifolds, which is Proposition 4.3 in Chapter 8 of do Carmo's book Riemannian Geometry. (Computing the sectional curvature of the tractricoid is relatively elementary, but I assume you already believe it has constant negative sectional curvature.)

In particular, any two nonsingular points $p$ and $q$ have neighborhoods $U$ and $V$ that are isometric; thus there is some $R > 0$ such that the metric discs $B_R(p)$ and $B_R(q)$ about these two points are isometric, and it follows (since volume and length are isometry invariants) that two metric discs about these points of the same radius $r < R$ will have equal areas (and similarly, metric circles will have equal circumferences).

However, for the tractricoid this is only a local result. For instance, if we imagine a sequence of points $p_n$ going off to infinity along a cusp of the tractricoid, then for any $\epsilon > 0$ there will be a point $p_n$ such that the disc of radius $\epsilon$ about $p_n$ overlaps itself (by wrapping around the cusp). Once this starts happening the area of this disc will be smaller than the area of a disc of radius $\epsilon$ centered at a point closer to the equator of the tractricoid where there is no overlap.

In summary: The hyperbolic plane is homogeneous, like the sphere, so areas of metric discs don't depend on basepoint. The tractricoid is not, and so you only get a result like this locally, the problem being when discs either run into the singular locus or when they overlap themselves from wrapping around the cusp.